0=4x^2+9x-28

Solution for 0=4x^2+9x-28 equation:

0=4x^2+9x-28

We move all terms to the left:

0-(4x^2+9x-28)=0

We add all the numbers together, and all the variables

-(4x^2+9x-28)=0

We get rid of parentheses

-4x^2-9x+28=0

a = -4; b = -9; c = +28;
Δ = b2-4ac
Δ = -92-4·(-4)·28
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-23}{2*-4}=\frac{-14}{-8}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+23}{2*-4}=\frac{32}{-8}
=-4
$