2x^2+6+7=303

Solution for 2x^2+6+7=303 equation:

2x^2+6+7=303

We move all terms to the left:

2x^2+6+7-(303)=0

We add all the numbers together, and all the variables

2x^2-290=0

a = 2; b = 0; c = -290;
Δ = b2-4ac
Δ = 02-4·2·(-290)
Δ = 2320
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:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{145}}{2*2}=\frac{0-4\sqrt{145}}{4}
=-\frac{4\sqrt{145}}{4}
=-\sqrt{145}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{145}}{2*2}=\frac{0+4\sqrt{145}}{4}
=\frac{4\sqrt{145}}{4}
=\sqrt{145}
$