3x^2+16x=5

Solution for 3x^2+16x=5 equation:

3x^2+16x=5

We move all terms to the left:

3x^2+16x-(5)=0

a = 3; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·3·(-5)
Δ = 316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{79}}{2*3}=\frac{-16-2\sqrt{79}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{79}}{2*3}=\frac{-16+2\sqrt{79}}{6}
$