H(x)=-16x^2+28x+3

Solution for H(x)=-16x^2+28x+3 equation:

(H)=-16H^2+28H+3

We move all terms to the left:

(H)-(-16H^2+28H+3)=0

We get rid of parentheses

16H^2-28H+H-3=0

We add all the numbers together, and all the variables

16H^2-27H-3=0

a = 16; b = -27; c = -3;
Δ = b2-4ac
Δ = -272-4·16·(-3)
Δ = 921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{921}}{2*16}=\frac{27-\sqrt{921}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{921}}{2*16}=\frac{27+\sqrt{921}}{32}
$