3x2+(4x-4)=3

Solution for 3x2+(4x-4)=3 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2+4x-7=0

a = 3; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·3·(-7)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10}{2*3}=\frac{-14}{6}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*3}=\frac{6}{6}
=1
$