4x^2-23x=-28

Solution for 4x^2-23x=-28 equation:

4x^2-23x=-28

We move all terms to the left:

4x^2-23x-(-28)=0

We add all the numbers together, and all the variables

4x^2-23x+28=0

a = 4; b = -23; c = +28;
Δ = b2-4ac
Δ = -232-4·4·28
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-9}{2*4}=\frac{14}{8}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+9}{2*4}=\frac{32}{8}
=4
$