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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -(4x+(5x-3)x), so:

4x+(5x-3)x

We multiply parentheses

5x^2+4x-3x

We add all the numbers together, and all the variables

5x^2+x

Back to the equation:

-(5x^2+x)



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5x^2+5=0

a = -5; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-5)·5
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-5}=\frac{-10}{-10}
=1
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-5}=\frac{10}{-10}
=-1
$

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