2x(3x-7)-5=(4x+5)

Solution for 2x(3x-7)-5=(4x+5) equation:

2x(3x-7)-5=(4x+5)

We move all terms to the left:

2x(3x-7)-5-((4x+5))=0

We multiply parentheses

6x^2-14x-((4x+5))-5=0


We calculate terms in parentheses: -((4x+5)), so:

(4x+5)

We get rid of parentheses

4x+5

Back to the equation:

-(4x+5)


We get rid of parentheses

6x^2-14x-4x-5-5=0

We add all the numbers together, and all the variables

6x^2-18x-10=0

a = 6; b = -18; c = -10;
Δ = b2-4ac
Δ = -182-4·6·(-10)
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{141}}{2*6}=\frac{18-2\sqrt{141}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{141}}{2*6}=\frac{18+2\sqrt{141}}{12}
$