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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2F^2+5F+12=0

a = -2; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-2)·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-2}=\frac{-16}{-4}
=+4
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-2}=\frac{6}{-4}
=-1+1/2
$