-3x^2+16x+35=0

Solution for -3x^2+16x+35=0 equation:

-3x^2+16x+35=0

a = -3; b = 16; c = +35;
Δ = b2-4ac
Δ = 162-4·(-3)·35
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-26}{2*-3}=\frac{-42}{-6}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+26}{2*-3}=\frac{10}{-6}
=-1+2/3
$