F(x)=5x^2+26x+5

Solution for F(x)=5x^2+26x+5 equation:

(F)=5F^2+26F+5

We move all terms to the left:

(F)-(5F^2+26F+5)=0

We get rid of parentheses

-5F^2+F-26F-5=0

We add all the numbers together, and all the variables

-5F^2-25F-5=0

a = -5; b = -25; c = -5;
Δ = b2-4ac
Δ = -252-4·(-5)·(-5)
Δ = 525
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{525}=\sqrt{25*21}=\sqrt{25}*\sqrt{21}=5\sqrt{21}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{21}}{2*-5}=\frac{25-5\sqrt{21}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{21}}{2*-5}=\frac{25+5\sqrt{21}}{-10}
$