(4x^2-5x-2)+(3x+x^2-8)=

Solution for (4x^2-5x-2)+(3x+x^2-8)= equation:

(4x^2-5x-2)+(3x+x^2-8)=

We move all terms to the left:

(4x^2-5x-2)+(3x+x^2-8)-()=0

We add all the numbers together, and all the variables

(3x+x^2-8)+(4x^2-5x-2)=0

We get rid of parentheses

x^2+4x^2+3x-5x-8-2=0

We add all the numbers together, and all the variables

5x^2-2x-10=0

a = 5; b = -2; c = -10;
Δ = b2-4ac
Δ = -22-4·5·(-10)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{51}}{2*5}=\frac{2-2\sqrt{51}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{51}}{2*5}=\frac{2+2\sqrt{51}}{10}
$