8x^2-8=7x^-2x

Solution for 8x^2-8=7x^-2x equation:

8x^2-8=7x^-2x

We move all terms to the left:

8x^2-8-(7x^-2x)=0

We get rid of parentheses

8x^2-7x^+2x-8=0

We add all the numbers together, and all the variables

8x^2-5x-8=0

a = 8; b = -5; c = -8;
Δ = b2-4ac
Δ = -52-4·8·(-8)
Δ = 281
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{281}}{2*8}=\frac{5-\sqrt{281}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{281}}{2*8}=\frac{5+\sqrt{281}}{16}
$