How do you find the roots of $2 m^{2} - 7 m - 13 = - 10$ $2m^2-7m-13=-10$ using the quadratic formula?

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Answer 1 · 2017-08-08T09:55:46

$\frac{7 \pm \sqrt{73}}{- 4}$ $(7+-sqrt73)/-4$

First, we must make the equation into the form of

$a x^{2} + b x + c$ $ax^2+bx+c$

in order to use the quadratic formula. Adding 10 to both sides of the equation gives us

$2 m^{2} - 7 m - 3$ $2m^2-7m-3$

This equation is in the form mentioned above and thus we can now use the quadratic equation to the roots

$a = 2, b = - 7, c = - 3$ $a= 2, b=-7, c=-3$

We substitute these value into the equation to give us

$\frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (2) (- 3)}}{- 2 \cdot (2)}$ $(-(-7)+-sqrt((-7)^2-4(2)(-3)))/(-2*(2))$

Thus, the roots to the equation $2 m^{2} - 7 m - 13 = - 10$ $2m^2-7m-13=-10$ would be

$\frac{7 \pm \sqrt{73}}{- 4}$ $(7+-sqrt73)/-4$

How do you find the roots of 2m2−7m−13=−102m^2-7m-13=-10 using the quadratic formula?