3(x^2+4x-7)=0

Solution for 3(x^2+4x-7)=0 equation:

3(x^2+4x-7)=0

We multiply parentheses

3x^2+12x-21=0

a = 3; b = 12; c = -21;
Δ = b2-4ac
Δ = 122-4·3·(-21)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{11}}{2*3}=\frac{-12-6\sqrt{11}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{11}}{2*3}=\frac{-12+6\sqrt{11}}{6}
$