every equations of the kind ax^2+bx+c=0 deserve to be solved using the quadratic formula: frac-b±sqrtb^2-4ac2a. The quadratic formula offers two solutions, one as soon as ± is addition and one as soon as it is subtraction.

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This equation is in conventional form: ax^2+bx+c=0. Instead of 1 for a, 6 for b, and -10-y because that c in the quadratic formula, frac-b±sqrtb^2-4ac2a. peak is atdisplaystyleleft(-3,-25 ight) Explanation: displaystyley=x^2+6x-16=x^2+6x+3^2-9-16=left(x+3 ight)^2-25 Comparing with ...
how do you resolve for displaystylex in the device of equations displaystyley=x^2+6x-17 and displaystyley=3x-7 ?
https://socratic.org/questions/how-do-you-solve-for-x-in-the-system-of-equations-y-x-2-6x-17-and-y-3x-7
x= 2 , x = - 5Explanation:Given 2 worths for y that are equal , thendisplaystylex^2+6x-17=3x-7 This is a quadratic equation , so collect terms to left next andequate to ...
bp Apr 17, 2015 A quadratic role represents a parabola, therefore its vertex and also the axis of symmetry have the right to be discovered from the provided equation as followsy=displaystyle3left(x^2+2x ight)-1 ...
-0.805 or 0.138Explanation:Assumingdisplaystyley=0you deserve to use the quadratic formuladisplaystylex=frac-b±sqrtb^2-4ac2a to discover the root of a ...
placed the equation right into vertex form to uncover that the vertex is atdisplaystyleleft(-8,-65 ight) Explanation:The vertex kind of a quadratic equation is displaystyley=aleft(x-h ight)^2+k ...
displaystyley=x^2+3x-10=left(x+5 ight)left(x-2 ight) Explanation:Find a pair of determinants ofdisplaystyle10which different bydisplaystyle3 .The pairdisplaystyle5,2 ...
More Items     All equations the the kind ax^2+bx+c=0 have the right to be resolved using the quadratic formula: frac-b±sqrtb^2-4ac2a. The quadratic formula offers two solutions, one once ± is enhancement and one once it is subtraction.
This equation is in conventional form: ax^2+bx+c=0. Substitute 1 for a, 6 for b, and -10-y for c in the quadratic formula, frac-b±sqrtb^2-4ac2a.
Divide 6, the coefficient that the x term, by 2 to acquire 3. Then include the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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Factor x^2+6x+9. In general, when x^2+bx+c is a perfect square, it can always be factored as left(x+fracb2 ight)^2.
left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray together l together 2 & 0 & 3 \ -1 & 1 & 5 endarray ight>    EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు