4a^2+25a=21

Solution for 4a^2+25a=21 equation:

4a^2+25a=21

We move all terms to the left:

4a^2+25a-(21)=0

a = 4; b = 25; c = -21;
Δ = b2-4ac
Δ = 252-4·4·(-21)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-31}{2*4}=\frac{-56}{8}
=-7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+31}{2*4}=\frac{6}{8}
=3/4
$