7r^2+54r=16

Solution for 7r^2+54r=16 equation:

7r^2+54r=16

We move all terms to the left:

7r^2+54r-(16)=0

a = 7; b = 54; c = -16;
Δ = b2-4ac
Δ = 542-4·7·(-16)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-58}{2*7}=\frac{-112}{14}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+58}{2*7}=\frac{4}{14}
=2/7
$