21x^2+28x-97=8

Solution for 21x^2+28x-97=8 equation:

21x^2+28x-97=8

We move all terms to the left:

21x^2+28x-97-(8)=0

We add all the numbers together, and all the variables

21x^2+28x-105=0

a = 21; b = 28; c = -105;
Δ = b2-4ac
Δ = 282-4·21·(-105)
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-98}{2*21}=\frac{-126}{42}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+98}{2*21}=\frac{70}{42}
=1+2/3
$