2x^2+7x+3=63

Solution for 2x^2+7x+3=63 equation:

2x^2+7x+3=63

We move all terms to the left:

2x^2+7x+3-(63)=0

We add all the numbers together, and all the variables

2x^2+7x-60=0

a = 2; b = 7; c = -60;
Δ = b2-4ac
Δ = 72-4·2·(-60)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*2}=\frac{-30}{4}
=-7+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*2}=\frac{16}{4}
=4
$