(4x-10)=(x+9)(x-5)

Solution for (4x-10)=(x+9)(x-5) equation:

(4x-10)=(x+9)(x-5)

We move all terms to the left:

(4x-10)-((x+9)(x-5))=0

We get rid of parentheses

4x-((x+9)(x-5))-10=0

We multiply parentheses ..

-((+x^2-5x+9x-45))+4x-10=0


We calculate terms in parentheses: -((+x^2-5x+9x-45)), so:

(+x^2-5x+9x-45)

We get rid of parentheses

x^2-5x+9x-45

We add all the numbers together, and all the variables

x^2+4x-45

Back to the equation:

-(x^2+4x-45)


We add all the numbers together, and all the variables

4x-(x^2+4x-45)-10=0

We get rid of parentheses

-x^2+4x-4x+45-10=0

We add all the numbers together, and all the variables

-1x^2+35=0

a = -1; b = 0; c = +35;
Δ = b2-4ac
Δ = 02-4·(-1)·35
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35}}{2*-1}=\frac{0-2\sqrt{35}}{-2}
=-\frac{2\sqrt{35}}{-2}
=-\frac{\sqrt{35}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35}}{2*-1}=\frac{0+2\sqrt{35}}{-2}
=\frac{2\sqrt{35}}{-2}
=\frac{\sqrt{35}}{-1}
$