(x^2+5x+4)-(2x^2-3x+6)=0

Solution for (x^2+5x+4)-(2x^2-3x+6)=0 equation:

(x^2+5x+4)-(2x^2-3x+6)=0

We get rid of parentheses

x^2-2x^2+5x+3x+4-6=0

We add all the numbers together, and all the variables

-1x^2+8x-2=0

a = -1; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·(-1)·(-2)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{14}}{2*-1}=\frac{-8-2\sqrt{14}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{14}}{2*-1}=\frac{-8+2\sqrt{14}}{-2}
$