10x^2+12x+4=20

Solution for 10x^2+12x+4=20 equation:

10x^2+12x+4=20

We move all terms to the left:

10x^2+12x+4-(20)=0

We add all the numbers together, and all the variables

10x^2+12x-16=0

a = 10; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·10·(-16)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*10}=\frac{-40}{20}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*10}=\frac{16}{20}
=4/5
$