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

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

X=(5X+2)(4X-3)

We move all terms to the left:

X-((5X+2)(4X-3))=0

We multiply parentheses ..

-((+20X^2-15X+8X-6))+X=0


We calculate terms in parentheses: -((+20X^2-15X+8X-6)), so:

(+20X^2-15X+8X-6)

We get rid of parentheses

20X^2-15X+8X-6

We add all the numbers together, and all the variables

20X^2-7X-6

Back to the equation:

-(20X^2-7X-6)


We add all the numbers together, and all the variables

X-(20X^2-7X-6)=0

We get rid of parentheses

-20X^2+X+7X+6=0

We add all the numbers together, and all the variables

-20X^2+8X+6=0

a = -20; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·(-20)·6
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{34}}{2*-20}=\frac{-8-4\sqrt{34}}{-40}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{34}}{2*-20}=\frac{-8+4\sqrt{34}}{-40}
$