F(-3x+7)=2x^2-5

Solution for F(-3x+7)=2x^2-5 equation:

(-3F+7)=2F^2-5

We move all terms to the left:

(-3F+7)-(2F^2-5)=0

We get rid of parentheses

-2F^2-3F+7+5=0

We add all the numbers together, and all the variables

-2F^2-3F+12=0

a = -2; b = -3; c = +12;
Δ = b2-4ac
Δ = -32-4·(-2)·12
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{105}}{2*-2}=\frac{3-\sqrt{105}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{105}}{2*-2}=\frac{3+\sqrt{105}}{-4}
$