10=-16x^2+48x+170

Solution for 10=-16x^2+48x+170 equation:

10=-16x^2+48x+170

We move all terms to the left:

10-(-16x^2+48x+170)=0

We get rid of parentheses

16x^2-48x-170+10=0

We add all the numbers together, and all the variables

16x^2-48x-160=0

a = 16; b = -48; c = -160;
Δ = b2-4ac
Δ = -482-4·16·(-160)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-112}{2*16}=\frac{-64}{32}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+112}{2*16}=\frac{160}{32}
=5
$