(x^2-3x+5)+(2x^2+11x-1)=0

Solution for (x^2-3x+5)+(2x^2+11x-1)=0 equation:

(x^2-3x+5)+(2x^2+11x-1)=0

We get rid of parentheses

x^2+2x^2-3x+11x+5-1=0

We add all the numbers together, and all the variables

3x^2+8x+4=0

a = 3; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·3·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*3}=\frac{-4}{6}
=-2/3
$