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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

10x^2-2x+5x-((3x-4))-15=0


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

(3x-4)

We get rid of parentheses

3x-4

Back to the equation:

-(3x-4)


We add all the numbers together, and all the variables

10x^2+3x-(3x-4)-15=0

We get rid of parentheses

10x^2+3x-3x+4-15=0

We add all the numbers together, and all the variables

10x^2-11=0

a = 10; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·10·(-11)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{110}}{2*10}=\frac{0-2\sqrt{110}}{20}
=-\frac{2\sqrt{110}}{20}
=-\frac{\sqrt{110}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{110}}{2*10}=\frac{0+2\sqrt{110}}{20}
=\frac{2\sqrt{110}}{20}
=\frac{\sqrt{110}}{10}
$