2x^2+3x-5=22

Solution for 2x^2+3x-5=22 equation:

2x^2+3x-5=22

We move all terms to the left:

2x^2+3x-5-(22)=0

We add all the numbers together, and all the variables

2x^2+3x-27=0

a = 2; b = 3; c = -27;
Δ = b2-4ac
Δ = 32-4·2·(-27)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*2}=\frac{-18}{4}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*2}=\frac{12}{4}
=3
$