(X^2+2x-3)+(x^2+3x-8)=180

Solution for (X^2+2x-3)+(x^2+3x-8)=180 equation:

(X^2+2X-3)+(X^2+3X-8)=180

We move all terms to the left:

(X^2+2X-3)+(X^2+3X-8)-(180)=0

We get rid of parentheses

X^2+X^2+2X+3X-3-8-180=0

We add all the numbers together, and all the variables

2X^2+5X-191=0

a = 2; b = 5; c = -191;
Δ = b2-4ac
Δ = 52-4·2·(-191)
Δ = 1553
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{-(5)-\sqrt{1553}}{2*2}=\frac{-5-\sqrt{1553}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1553}}{2*2}=\frac{-5+\sqrt{1553}}{4}
$