0=-16x^2+95x-98

Solution for 0=-16x^2+95x-98 equation:

0=-16x^2+95x-98

We move all terms to the left:

0-(-16x^2+95x-98)=0

We add all the numbers together, and all the variables

-(-16x^2+95x-98)=0

We get rid of parentheses

16x^2-95x+98=0

a = 16; b = -95; c = +98;
Δ = b2-4ac
Δ = -952-4·16·98
Δ = 2753
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{-(-95)-\sqrt{2753}}{2*16}=\frac{95-\sqrt{2753}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{2753}}{2*16}=\frac{95+\sqrt{2753}}{32}
$