(x^2+5x-80)+(4x+10)=x

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Solution for (x^2+5x-80)+(4x+10)=x equation:



(x^2+5x-80)+(4x+10)=x
We move all terms to the left:
(x^2+5x-80)+(4x+10)-(x)=0
We add all the numbers together, and all the variables
-1x+(x^2+5x-80)+(4x+10)=0
We get rid of parentheses
x^2-1x+5x+4x-80+10=0
We add all the numbers together, and all the variables
x^2+8x-70=0
a = 1; b = 8; c = -70;
Δ = b2-4ac
Δ = 82-4·1·(-70)
Δ = 344
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:
$\sqrt{\Delta}=\sqrt{344}=\sqrt{4*86}=\sqrt{4}*\sqrt{86}=2\sqrt{86}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{86}}{2*1}=\frac{-8-2\sqrt{86}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{86}}{2*1}=\frac{-8+2\sqrt{86}}{2} $

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