4x^2+x-20=-6

Solution for 4x^2+x-20=-6 equation:

4x^2+x-20=-6

We move all terms to the left:

4x^2+x-20-(-6)=0

We add all the numbers together, and all the variables

4x^2+x-14=0

a = 4; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·4·(-14)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*4}=\frac{-16}{8}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*4}=\frac{14}{8}
=1+3/4
$