7x^2=21x+3=x^2+8x-6

Solution for 7x^2=21x+3=x^2+8x-6 equation:

7x^2=21x+3=x^2+8x-6

We move all terms to the left:

7x^2-(21x+3)=0

We get rid of parentheses

7x^2-21x-3=0

a = 7; b = -21; c = -3;
Δ = b2-4ac
Δ = -212-4·7·(-3)
Δ = 525
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{525}=\sqrt{25*21}=\sqrt{25}*\sqrt{21}=5\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-5\sqrt{21}}{2*7}=\frac{21-5\sqrt{21}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+5\sqrt{21}}{2*7}=\frac{21+5\sqrt{21}}{14}
$