-4c2+19c=21

Solution for -4c2+19c=21 equation:

-4c^2+19c=21

We move all terms to the left:

-4c^2+19c-(21)=0

a = -4; b = 19; c = -21;
Δ = b2-4ac
Δ = 192-4·(-4)·(-21)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*-4}=\frac{-24}{-8}
=+3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*-4}=\frac{-14}{-8}
=1+3/4
$