2x^2=10*x+8

Solution for 2x^2=10*x+8 equation:

2x^2=10x+8

We move all terms to the left:

2x^2-(10x+8)=0

We get rid of parentheses

2x^2-10x-8=0

a = 2; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·2·(-8)
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{41}}{2*2}=\frac{10-2\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{41}}{2*2}=\frac{10+2\sqrt{41}}{4}
$