3x^2=-5+16x

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

3x^2=-5+16x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

3x^2-16x+5=0

a = 3; b = -16; c = +5;
Δ = b2-4ac
Δ = -162-4·3·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*3}=\frac{2}{6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*3}=\frac{30}{6}
=5
$