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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+20X^2-12X-15X+9))+X=0


We calculate terms in parentheses: -((+20X^2-12X-15X+9)), so:

(+20X^2-12X-15X+9)

We get rid of parentheses

20X^2-12X-15X+9

We add all the numbers together, and all the variables

20X^2-27X+9

Back to the equation:

-(20X^2-27X+9)


We add all the numbers together, and all the variables

X-(20X^2-27X+9)=0

We get rid of parentheses

-20X^2+X+27X-9=0

We add all the numbers together, and all the variables

-20X^2+28X-9=0

a = -20; b = 28; c = -9;
Δ = b2-4ac
Δ = 282-4·(-20)·(-9)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*-20}=\frac{-36}{-40}
=9/10
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*-20}=\frac{-20}{-40}
=1/2
$