(2x^2)+(3x-5)=90

Solution for (2x^2)+(3x-5)=90 equation:

(2x^2)+(3x-5)=90

We move all terms to the left:

(2x^2)+(3x-5)-(90)=0

We get rid of parentheses

2x^2+3x-5-90=0

We add all the numbers together, and all the variables

2x^2+3x-95=0

a = 2; b = 3; c = -95;
Δ = b2-4ac
Δ = 32-4·2·(-95)
Δ = 769
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{769}}{2*2}=\frac{-3-\sqrt{769}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{769}}{2*2}=\frac{-3+\sqrt{769}}{4}
$