(2x^2+3)+(5x-10)+20=50

Solution for (2x^2+3)+(5x-10)+20=50 equation:

(2x^2+3)+(5x-10)+20=50

We move all terms to the left:

(2x^2+3)+(5x-10)+20-(50)=0

We add all the numbers together, and all the variables

(2x^2+3)+(5x-10)-30=0

We get rid of parentheses

2x^2+5x+3-10-30=0

We add all the numbers together, and all the variables

2x^2+5x-37=0

a = 2; b = 5; c = -37;
Δ = b2-4ac
Δ = 52-4·2·(-37)
Δ = 321
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{321}}{2*2}=\frac{-5-\sqrt{321}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{321}}{2*2}=\frac{-5+\sqrt{321}}{4}
$