2x^2+4x+49=100

Solution for 2x^2+4x+49=100 equation:

2x^2+4x+49=100

We move all terms to the left:

2x^2+4x+49-(100)=0

We add all the numbers together, and all the variables

2x^2+4x-51=0

a = 2; b = 4; c = -51;
Δ = b2-4ac
Δ = 42-4·2·(-51)
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{106}}{2*2}=\frac{-4-2\sqrt{106}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{106}}{2*2}=\frac{-4+2\sqrt{106}}{4}
$