35=-16x^2+72

Solution for 35=-16x^2+72 equation:

35=-16x^2+72

We move all terms to the left:

35-(-16x^2+72)=0

We get rid of parentheses

16x^2-72+35=0

We add all the numbers together, and all the variables

16x^2-37=0

a = 16; b = 0; c = -37;
Δ = b2-4ac
Δ = 02-4·16·(-37)
Δ = 2368
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2368}=\sqrt{64*37}=\sqrt{64}*\sqrt{37}=8\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{37}}{2*16}=\frac{0-8\sqrt{37}}{32}
=-\frac{8\sqrt{37}}{32}
=-\frac{\sqrt{37}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{37}}{2*16}=\frac{0+8\sqrt{37}}{32}
=\frac{8\sqrt{37}}{32}
=\frac{\sqrt{37}}{4}
$