3x(2x+50)=1700

Solution for 3x(2x+50)=1700 equation:

3x(2x+50)=1700

We move all terms to the left:

3x(2x+50)-(1700)=0

We multiply parentheses

6x^2+150x-1700=0

a = 6; b = 150; c = -1700;
Δ = b2-4ac
Δ = 1502-4·6·(-1700)
Δ = 63300
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:

$\sqrt{\Delta}=\sqrt{63300}=\sqrt{100*633}=\sqrt{100}*\sqrt{633}=10\sqrt{633}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-10\sqrt{633}}{2*6}=\frac{-150-10\sqrt{633}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+10\sqrt{633}}{2*6}=\frac{-150+10\sqrt{633}}{12}
$