7w^2+42=98

Solution for 7w^2+42=98 equation:

7w^2+42=98

We move all terms to the left:

7w^2+42-(98)=0

We add all the numbers together, and all the variables

7w^2-56=0

a = 7; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·7·(-56)
Δ = 1568
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{2}}{2*7}=\frac{0-28\sqrt{2}}{14}
=-\frac{28\sqrt{2}}{14}
=-2\sqrt{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{2}}{2*7}=\frac{0+28\sqrt{2}}{14}
=\frac{28\sqrt{2}}{14}
=2\sqrt{2}
$