(3x^2-14x+16)=0

Solution for (3x^2-14x+16)=0 equation:

(3x^2-14x+16)=0

We get rid of parentheses

3x^2-14x+16=0

a = 3; b = -14; c = +16;
Δ = b2-4ac
Δ = -142-4·3·16
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*3}=\frac{12}{6}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*3}=\frac{16}{6}
=2+2/3
$