8-5(3-2x)=(x-4)(x+9)

Solution for 8-5(3-2x)=(x-4)(x+9) equation:

8-5(3-2x)=(x-4)(x+9)

We move all terms to the left:

8-5(3-2x)-((x-4)(x+9))=0

We add all the numbers together, and all the variables

-5(-2x+3)-((x-4)(x+9))+8=0

We multiply parentheses

10x-((x-4)(x+9))-15+8=0

We multiply parentheses ..

-((+x^2+9x-4x-36))+10x-15+8=0


We calculate terms in parentheses: -((+x^2+9x-4x-36)), so:

(+x^2+9x-4x-36)

We get rid of parentheses

x^2+9x-4x-36

We add all the numbers together, and all the variables

x^2+5x-36

Back to the equation:

-(x^2+5x-36)


We add all the numbers together, and all the variables

10x-(x^2+5x-36)-7=0

We get rid of parentheses

-x^2+10x-5x+36-7=0

We add all the numbers together, and all the variables

-1x^2+5x+29=0

a = -1; b = 5; c = +29;
Δ = b2-4ac
Δ = 52-4·(-1)·29
Δ = 141
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{-(5)-\sqrt{141}}{2*-1}=\frac{-5-\sqrt{141}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{141}}{2*-1}=\frac{-5+\sqrt{141}}{-2}
$