3x2+28x+48=0

Solution for 3x2+28x+48=0 equation:

3x^2+28x+48=0

a = 3; b = 28; c = +48;
Δ = b2-4ac
Δ = 282-4·3·48
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{13}}{2*3}=\frac{-28-4\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{13}}{2*3}=\frac{-28+4\sqrt{13}}{6}
$