2x^2+6x+7=303

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

2x^2+6x+7=303

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+6x-296=0

a = 2; b = 6; c = -296;
Δ = b2-4ac
Δ = 62-4·2·(-296)
Δ = 2404
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:

$\sqrt{\Delta}=\sqrt{2404}=\sqrt{4*601}=\sqrt{4}*\sqrt{601}=2\sqrt{601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{601}}{2*2}=\frac{-6-2\sqrt{601}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{601}}{2*2}=\frac{-6+2\sqrt{601}}{4}
$