8x^2-5=-39x

Solution for 8x^2-5=-39x equation:

8x^2-5=-39x

We move all terms to the left:

8x^2-5-(-39x)=0

We get rid of parentheses

8x^2+39x-5=0

a = 8; b = 39; c = -5;
Δ = b2-4ac
Δ = 392-4·8·(-5)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-41}{2*8}=\frac{-80}{16}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+41}{2*8}=\frac{2}{16}
=1/8
$