3x^2=2(x-1)+7

Solution for 3x^2=2(x-1)+7 equation:

3x^2=2(x-1)+7

We move all terms to the left:

3x^2-(2(x-1)+7)=0


We calculate terms in parentheses: -(2(x-1)+7), so:

2(x-1)+7

We multiply parentheses

2x-2+7

We add all the numbers together, and all the variables

2x+5

Back to the equation:

-(2x+5)


We get rid of parentheses

3x^2-2x-5=0

a = 3; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·3·(-5)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*3}=\frac{-6}{6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*3}=\frac{10}{6}
=1+2/3
$