x^2+7x-33=3(x-4)

Solution for x^2+7x-33=3(x-4) equation:

x^2+7x-33=3(x-4)

We move all terms to the left:

x^2+7x-33-(3(x-4))=0


We calculate terms in parentheses: -(3(x-4)), so:

3(x-4)

We multiply parentheses

3x-12

Back to the equation:

-(3x-12)


We get rid of parentheses

x^2+7x-3x+12-33=0

We add all the numbers together, and all the variables

x^2+4x-21=0

a = 1; b = 4; c = -21;
Δ = b2-4ac
Δ = 42-4·1·(-21)
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*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*1}=\frac{6}{2}
=3
$