F(x)=3x2+4x+-7

Solution for F(x)=3x2+4x+-7 equation:

(F)=3F^2+4F+-7

We move all terms to the left:

(F)-(3F^2+4F+-7)=0

We use the square of the difference formula

F-(3F^2+4F-7)=0

We get rid of parentheses

-3F^2+F-4F+7=0

We add all the numbers together, and all the variables

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

a = -3; b = -3; c = +7;
Δ = b2-4ac
Δ = -32-4·(-3)·7
Δ = 93
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{93}}{2*-3}=\frac{3-\sqrt{93}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{93}}{2*-3}=\frac{3+\sqrt{93}}{-6}
$