16x^2-49x-225=0

Solution for 16x^2-49x-225=0 equation:

16x^2-49x-225=0

a = 16; b = -49; c = -225;
Δ = b2-4ac
Δ = -492-4·16·(-225)
Δ = 16801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-\sqrt{16801}}{2*16}=\frac{49-\sqrt{16801}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{16801}}{2*16}=\frac{49+\sqrt{16801}}{32}
$