2x(x+5)-3(x^2+2x-1)=9-5x-x^

Solution for 2x(x+5)-3(x^2+2x-1)=9-5x-x^ equation:

2x(x+5)-3(x^2+2x-1)=9-5x-x^

We move all terms to the left:

2x(x+5)-3(x^2+2x-1)-(9-5x-x^)=0

We multiply parentheses

2x^2-3x^2+10x-6x-(9-5x-x^)+3=0

We get rid of parentheses

2x^2-3x^2+10x-6x+5x+x^-9+3=0

We add all the numbers together, and all the variables

-1x^2+10x-6=0

a = -1; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·(-1)·(-6)
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{-(10)-2\sqrt{19}}{2*-1}=\frac{-10-2\sqrt{19}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{19}}{2*-1}=\frac{-10+2\sqrt{19}}{-2}
$