10=(4x)(6x-58)

Solution for 10=(4x)(6x-58) equation:

10=(4x)(6x-58)

We move all terms to the left:

10-((4x)(6x-58))=0


We calculate terms in parentheses: -(4x(6x-58)), so:

4x(6x-58)

We multiply parentheses

24x^2-232x

Back to the equation:

-(24x^2-232x)


We get rid of parentheses

-24x^2+232x+10=0

a = -24; b = 232; c = +10;
Δ = b2-4ac
Δ = 2322-4·(-24)·10
Δ = 54784
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:

$\sqrt{\Delta}=\sqrt{54784}=\sqrt{256*214}=\sqrt{256}*\sqrt{214}=16\sqrt{214}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(232)-16\sqrt{214}}{2*-24}=\frac{-232-16\sqrt{214}}{-48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(232)+16\sqrt{214}}{2*-24}=\frac{-232+16\sqrt{214}}{-48}
$