35q^2+34q-21=0

Solution for 35q^2+34q-21=0 equation:

35q^2+34q-21=0

a = 35; b = 34; c = -21;
Δ = b2-4ac
Δ = 342-4·35·(-21)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-64}{2*35}=\frac{-98}{70}
=-1+2/5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+64}{2*35}=\frac{30}{70}
=3/7
$