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

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

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

We move all terms to the left:

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


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

2(x+2)

We multiply parentheses

2x+4

Back to the equation:

-(2x+4)


We get rid of parentheses

5x^2-2x-4-4=0

We add all the numbers together, and all the variables

5x^2-2x-8=0

a = 5; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·5·(-8)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{41}}{2*5}=\frac{2-2\sqrt{41}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{41}}{2*5}=\frac{2+2\sqrt{41}}{10}
$