5(3x)^2-4(3x)=0

Solution for 5(3x)^2-4(3x)=0 equation:

5(3x)^2-4(3x)=0

a = 53; b = -43; c = 0;
Δ = b2-4ac
Δ = -432-4·53·0
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-43}{2*53}=\frac{0}{106}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+43}{2*53}=\frac{86}{106}
=43/53
$