(4x-5)=(2x^2-3x-3)

Solution for (4x-5)=(2x^2-3x-3) equation:

(4x-5)=(2x^2-3x-3)

We move all terms to the left:

(4x-5)-((2x^2-3x-3))=0

We get rid of parentheses

4x-((2x^2-3x-3))-5=0


We calculate terms in parentheses: -((2x^2-3x-3)), so:

(2x^2-3x-3)

We get rid of parentheses

2x^2-3x-3

Back to the equation:

-(2x^2-3x-3)


We get rid of parentheses

-2x^2+4x+3x+3-5=0

We add all the numbers together, and all the variables

-2x^2+7x-2=0

a = -2; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·(-2)·(-2)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{33}}{2*-2}=\frac{-7-\sqrt{33}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{33}}{2*-2}=\frac{-7+\sqrt{33}}{-4}
$