How do you solve $5 x^{2} + 5 = - 13 x$ $5x^2+5=-13x$ using the quadratic formula?

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Answer 1 · 2017-06-03T08:38:25

$x = - \frac{13}{10} + \sqrt{69}, or - \frac{13}{10} - \sqrt{69}$ $x=-13/10+sqrt(69), or -13/10-sqrt(69)$

Firstly, in any quadratic equation, you want to get it in the form $a x^{2} + b x + c$ $ax^2+bx+c$ .

To do that, add $13 x$ $13x$ to both sides.

$5 x^{2} + 13 x + 5 = 0$ $5x^2+13x+5=0$

Now, we can get the following ; $a = 5, b = 13, c = 5$ $a=5, b=13, c=5$

The quadratic formula is this : $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

Substitute the values

$\frac{- 13 \pm \sqrt{13^{2} - (4) (5) (5)}}{(2) (5)}$ $(-13+-sqrt(13^2-(4)(5)(5)))/((2)(5))$

$\frac{- 13 + \sqrt{69}}{10}$ $(-13+sqrt(69))/10$

$- \frac{13}{10} + \sqrt{69}, - \frac{13}{10} - \sqrt{69}$ $-13/10+sqrt(69), -13/10-sqrt(69)$

How do you solve 5x^2+5=-13x5x2+5=−13x5x^2+5=-13x using the quadratic formula?