21a^2+60a+17=0

Solution for 21a^2+60a+17=0 equation:

21a^2+60a+17=0

a = 21; b = 60; c = +17;
Δ = b2-4ac
Δ = 602-4·21·17
Δ = 2172
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2172}=\sqrt{4*543}=\sqrt{4}*\sqrt{543}=2\sqrt{543}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-2\sqrt{543}}{2*21}=\frac{-60-2\sqrt{543}}{42}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+2\sqrt{543}}{2*21}=\frac{-60+2\sqrt{543}}{42}
$