7x^2+13x-20=4

Solution for 7x^2+13x-20=4 equation:

7x^2+13x-20=4

We move all terms to the left:

7x^2+13x-20-(4)=0

We add all the numbers together, and all the variables

7x^2+13x-24=0

a = 7; b = 13; c = -24;
Δ = b2-4ac
Δ = 132-4·7·(-24)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-29}{2*7}=\frac{-42}{14}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+29}{2*7}=\frac{16}{14}
=1+1/7
$