28w2+13=20

Solution for 28w2+13=20 equation:

28w^2+13=20

We move all terms to the left:

28w^2+13-(20)=0

We add all the numbers together, and all the variables

28w^2-7=0

a = 28; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·28·(-7)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*28}=\frac{-28}{56}
=-1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*28}=\frac{28}{56}
=1/2
$