0=-2x^2+28x-40

Solution for 0=-2x^2+28x-40 equation:

0=-2x^2+28x-40

We move all terms to the left:

0-(-2x^2+28x-40)=0

We add all the numbers together, and all the variables

-(-2x^2+28x-40)=0

We get rid of parentheses

2x^2-28x+40=0

a = 2; b = -28; c = +40;
Δ = b2-4ac
Δ = -282-4·2·40
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{29}}{2*2}=\frac{28-4\sqrt{29}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{29}}{2*2}=\frac{28+4\sqrt{29}}{4}
$