4x^2-49+(4x-14)(x+3)=0

Solution for 4x^2-49+(4x-14)(x+3)=0 equation:

4x^2-49+(4x-14)(x+3)=0

We multiply parentheses ..

4x^2+(+4x^2+12x-14x-42)-49=0

We get rid of parentheses

4x^2+4x^2+12x-14x-42-49=0

We add all the numbers together, and all the variables

8x^2-2x-91=0

a = 8; b = -2; c = -91;
Δ = b2-4ac
Δ = -22-4·8·(-91)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-54}{2*8}=\frac{-52}{16}
=-3+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+54}{2*8}=\frac{56}{16}
=3+1/2
$