7x^2+21=196

Solution for 7x^2+21=196 equation:

7x^2+21=196

We move all terms to the left:

7x^2+21-(196)=0

We add all the numbers together, and all the variables

7x^2-175=0

a = 7; b = 0; c = -175;
Δ = b2-4ac
Δ = 02-4·7·(-175)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-70}{2*7}=\frac{-70}{14}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+70}{2*7}=\frac{70}{14}
=5
$