7x^2+10=51

Solution for 7x^2+10=51 equation:

7x^2+10=51

We move all terms to the left:

7x^2+10-(51)=0

We add all the numbers together, and all the variables

7x^2-41=0

a = 7; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·7·(-41)
Δ = 1148
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:

$\sqrt{\Delta}=\sqrt{1148}=\sqrt{4*287}=\sqrt{4}*\sqrt{287}=2\sqrt{287}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{287}}{2*7}=\frac{0-2\sqrt{287}}{14}
=-\frac{2\sqrt{287}}{14}
=-\frac{\sqrt{287}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{287}}{2*7}=\frac{0+2\sqrt{287}}{14}
=\frac{2\sqrt{287}}{14}
=\frac{\sqrt{287}}{7}
$