7x(4x-5)=21

Solution for 7x(4x-5)=21 equation:

7x(4x-5)=21

We move all terms to the left:

7x(4x-5)-(21)=0

We multiply parentheses

28x^2-35x-21=0

a = 28; b = -35; c = -21;
Δ = b2-4ac
Δ = -352-4·28·(-21)
Δ = 3577
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:

$\sqrt{\Delta}=\sqrt{3577}=\sqrt{49*73}=\sqrt{49}*\sqrt{73}=7\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-7\sqrt{73}}{2*28}=\frac{35-7\sqrt{73}}{56}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+7\sqrt{73}}{2*28}=\frac{35+7\sqrt{73}}{56}
$