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

Solution for (x^2+9x-3)=6x-5 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+3x+2=0

a = 1; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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