6x^2+7x-33=3(5-3x)

Solution for 6x^2+7x-33=3(5-3x) equation:

6x^2+7x-33=3(5-3x)

We move all terms to the left:

6x^2+7x-33-(3(5-3x))=0

We add all the numbers together, and all the variables

6x^2+7x-(3(-3x+5))-33=0


We calculate terms in parentheses: -(3(-3x+5)), so:

3(-3x+5)

We multiply parentheses

-9x+15

Back to the equation:

-(-9x+15)


We get rid of parentheses

6x^2+7x+9x-15-33=0

We add all the numbers together, and all the variables

6x^2+16x-48=0

a = 6; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·6·(-48)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{22}}{2*6}=\frac{-16-8\sqrt{22}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{22}}{2*6}=\frac{-16+8\sqrt{22}}{12}
$