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

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

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

We get rid of parentheses

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

We multiply parentheses ..

x^2+(+2x^2-10x+3x-15)+3x-1=0

We get rid of parentheses

x^2+2x^2-10x+3x+3x-15-1=0

We add all the numbers together, and all the variables

3x^2-4x-16=0

a = 3; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·3·(-16)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{13}}{2*3}=\frac{4-4\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{13}}{2*3}=\frac{4+4\sqrt{13}}{6}
$