0=-8x^2-25x+28

Solution for 0=-8x^2-25x+28 equation:

0=-8x^2-25x+28

We move all terms to the left:

0-(-8x^2-25x+28)=0

We add all the numbers together, and all the variables

-(-8x^2-25x+28)=0

We get rid of parentheses

8x^2+25x-28=0

a = 8; b = 25; c = -28;
Δ = b2-4ac
Δ = 252-4·8·(-28)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-39}{2*8}=\frac{-64}{16}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+39}{2*8}=\frac{14}{16}
=7/8
$