21x^2+16x-22=0

Solution for 21x^2+16x-22=0 equation:

21x^2+16x-22=0

a = 21; b = 16; c = -22;
Δ = b2-4ac
Δ = 162-4·21·(-22)
Δ = 2104
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:

$\sqrt{\Delta}=\sqrt{2104}=\sqrt{4*526}=\sqrt{4}*\sqrt{526}=2\sqrt{526}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{526}}{2*21}=\frac{-16-2\sqrt{526}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{526}}{2*21}=\frac{-16+2\sqrt{526}}{42}
$