(6x^2-x-4)+(2x^2+5x-5)=0

Solution for (6x^2-x-4)+(2x^2+5x-5)=0 equation:

(6x^2-x-4)+(2x^2+5x-5)=0

We get rid of parentheses

6x^2+2x^2-x+5x-4-5=0

We add all the numbers together, and all the variables

8x^2+4x-9=0

a = 8; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·8·(-9)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{19}}{2*8}=\frac{-4-4\sqrt{19}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{19}}{2*8}=\frac{-4+4\sqrt{19}}{16}
$