10(x^2-4)=5(x^2+1)

Solution for 10(x^2-4)=5(x^2+1) equation:

10(x^2-4)=5(x^2+1)

We move all terms to the left:

10(x^2-4)-(5(x^2+1))=0

We multiply parentheses

10x^2-(5(x^2+1))-40=0


We calculate terms in parentheses: -(5(x^2+1)), so:

5(x^2+1)

We multiply parentheses

5x^2+5

Back to the equation:

-(5x^2+5)


We get rid of parentheses

10x^2-5x^2-5-40=0

We add all the numbers together, and all the variables

5x^2-45=0

a = 5; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·5·(-45)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*5}=\frac{-30}{10}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*5}=\frac{30}{10}
=3
$