50=300-10(x^2+10)

Solution for 50=300-10(x^2+10) equation:

50=300-10(x^2+10)

We move all terms to the left:

50-(300-10(x^2+10))=0


We calculate terms in parentheses: -(300-10(x^2+10)), so:

300-10(x^2+10)

determiningTheFunctionDomain
-10(x^2+10)+300

We multiply parentheses

-10x^2-100+300

We add all the numbers together, and all the variables

-10x^2+200

Back to the equation:

-(-10x^2+200)


We get rid of parentheses

10x^2-200+50=0

We add all the numbers together, and all the variables

10x^2-150=0

a = 10; b = 0; c = -150;
Δ = b2-4ac
Δ = 02-4·10·(-150)
Δ = 6000
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:

$\sqrt{\Delta}=\sqrt{6000}=\sqrt{400*15}=\sqrt{400}*\sqrt{15}=20\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{15}}{2*10}=\frac{0-20\sqrt{15}}{20}
=-\frac{20\sqrt{15}}{20}
=-\sqrt{15}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{15}}{2*10}=\frac{0+20\sqrt{15}}{20}
=\frac{20\sqrt{15}}{20}
=\sqrt{15}
$