(6x+10)=(-x^2+50)

Solution for (6x+10)=(-x^2+50) equation:

(6x+10)=(-x^2+50)

We move all terms to the left:

(6x+10)-((-x^2+50))=0

We get rid of parentheses

-((-x^2+50))+6x+10=0


We calculate terms in parentheses: -((-x^2+50)), so:

(-x^2+50)

We get rid of parentheses

-x^2+50

We add all the numbers together, and all the variables

-1x^2+50

Back to the equation:

-(-1x^2+50)


We get rid of parentheses

1x^2+6x-50+10=0

We add all the numbers together, and all the variables

x^2+6x-40=0

a = 1; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·1·(-40)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*1}=\frac{8}{2}
=4
$