6x2+20=4(3x+5)

Solution for 6x2+20=4(3x+5) equation:

6x^2+20=4(3x+5)

We move all terms to the left:

6x^2+20-(4(3x+5))=0


We calculate terms in parentheses: -(4(3x+5)), so:

4(3x+5)

We multiply parentheses

12x+20

Back to the equation:

-(12x+20)


We get rid of parentheses

6x^2-12x-20+20=0

We add all the numbers together, and all the variables

6x^2-12x=0

a = 6; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·6·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*6}=\frac{0}{12}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*6}=\frac{24}{12}
=2
$