4x-3x^=(x+1)(x-5)

Solution for 4x-3x^=(x+1)(x-5) equation:

4x-3x^=(x+1)(x-5)

We move all terms to the left:

4x-3x^-((x+1)(x-5))=0

We add all the numbers together, and all the variables

x-((x+1)(x-5))=0

We multiply parentheses ..

-((+x^2-5x+x-5))+x=0


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

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

We get rid of parentheses

x^2-5x+x-5

We add all the numbers together, and all the variables

x^2-4x-5

Back to the equation:

-(x^2-4x-5)


We add all the numbers together, and all the variables

x-(x^2-4x-5)=0

We get rid of parentheses

-x^2+x+4x+5=0

We add all the numbers together, and all the variables

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

a = -1; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·(-1)·5
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{5}}{2*-1}=\frac{-5-3\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{5}}{2*-1}=\frac{-5+3\sqrt{5}}{-2}
$