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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8x^2-14x-15

Back to the equation:

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


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -8; b = 15; c = +15;
Δ = b2-4ac
Δ = 152-4·(-8)·15
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*-8}=\frac{-15-\sqrt{705}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{705}}{2*-8}=\frac{-15+\sqrt{705}}{-16}
$