r^2+6+7r-8=0

Solution for r^2+6+7r-8=0 equation:

r^2+6+7r-8=0

We add all the numbers together, and all the variables

r^2+7r-2=0

a = 1; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·1·(-2)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{57}}{2*1}=\frac{-7-\sqrt{57}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{57}}{2*1}=\frac{-7+\sqrt{57}}{2}
$