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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

15X^2-14X-8

Back to the equation:

-(15X^2-14X-8)


We add all the numbers together, and all the variables

X-(15X^2-14X-8)=0

We get rid of parentheses

-15X^2+X+14X+8=0

We add all the numbers together, and all the variables

-15X^2+15X+8=0

a = -15; b = 15; c = +8;
Δ = b2-4ac
Δ = 152-4·(-15)·8
Δ = 705
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{-(15)-\sqrt{705}}{2*-15}=\frac{-15-\sqrt{705}}{-30}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{705}}{2*-15}=\frac{-15+\sqrt{705}}{-30}
$