(2x^2+5x-3)+(3x+7)=0

Solution for (2x^2+5x-3)+(3x+7)=0 equation:

(2x^2+5x-3)+(3x+7)=0

We get rid of parentheses

2x^2+5x+3x-3+7=0

We add all the numbers together, and all the variables

2x^2+8x+4=0

a = 2; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·2·4
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2}}{2*2}=\frac{-8-4\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2}}{2*2}=\frac{-8+4\sqrt{2}}{4}
$