2x^2-3x-8=3(x-3)

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

2x^2-3x-8=3(x-3)

We move all terms to the left:

2x^2-3x-8-(3(x-3))=0


We calculate terms in parentheses: -(3(x-3)), so:

3(x-3)

We multiply parentheses

3x-9

Back to the equation:

-(3x-9)


We get rid of parentheses

2x^2-3x-3x+9-8=0

We add all the numbers together, and all the variables

2x^2-6x+1=0

a = 2; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·2·1
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{7}}{2*2}=\frac{6-2\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{7}}{2*2}=\frac{6+2\sqrt{7}}{4}
$