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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-1x-1=0

a = 2; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·2·(-1)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*2}=\frac{-2}{4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*2}=\frac{4}{4}
=1
$