10x^2-28x=16

Solution for 10x^2-28x=16 equation:

10x^2-28x=16

We move all terms to the left:

10x^2-28x-(16)=0

a = 10; b = -28; c = -16;
Δ = b2-4ac
Δ = -282-4·10·(-16)
Δ = 1424
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:

$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{89}}{2*10}=\frac{28-4\sqrt{89}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{89}}{2*10}=\frac{28+4\sqrt{89}}{20}
$