4r^2+3r-13=-6

Solution for 4r^2+3r-13=-6 equation:

4r^2+3r-13=-6

We move all terms to the left:

4r^2+3r-13-(-6)=0

We add all the numbers together, and all the variables

4r^2+3r-7=0

a = 4; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·4·(-7)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*4}=\frac{-14}{8}
=-1+3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*4}=\frac{8}{8}
=1
$