(x2-3x-3)+(2x2+7x-2)=0

Solution for (x2-3x-3)+(2x2+7x-2)=0 equation:

(x2-3x-3)+(2x^2+7x-2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2+4x-5=0

a = 3; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·3·(-5)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{19}}{2*3}=\frac{-4-2\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{19}}{2*3}=\frac{-4+2\sqrt{19}}{6}
$