3x^=3-30x^2+72x

Solution for 3x^=3-30x^2+72x equation:

3x^=3-30x^2+72x

We move all terms to the left:

3x^-(3-30x^2+72x)=0

We add all the numbers together, and all the variables

-(3-30x^2+72x)+3x=0

We get rid of parentheses

30x^2-72x+3x-3=0

We add all the numbers together, and all the variables

30x^2-69x-3=0

a = 30; b = -69; c = -3;
Δ = b2-4ac
Δ = -692-4·30·(-3)
Δ = 5121
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:

$\sqrt{\Delta}=\sqrt{5121}=\sqrt{9*569}=\sqrt{9}*\sqrt{569}=3\sqrt{569}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-69)-3\sqrt{569}}{2*30}=\frac{69-3\sqrt{569}}{60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-69)+3\sqrt{569}}{2*30}=\frac{69+3\sqrt{569}}{60}
$