F(x)=(-2x+10)(-2x+15)

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Solution for F(x)=(-2x+10)(-2x+15) equation:



(F)=(-2F+10)(-2F+15)
We move all terms to the left:
(F)-((-2F+10)(-2F+15))=0
We multiply parentheses ..
-((+4F^2-30F-20F+150))+F=0
We calculate terms in parentheses: -((+4F^2-30F-20F+150)), so:
(+4F^2-30F-20F+150)
We get rid of parentheses
4F^2-30F-20F+150
We add all the numbers together, and all the variables
4F^2-50F+150
Back to the equation:
-(4F^2-50F+150)
We add all the numbers together, and all the variables
F-(4F^2-50F+150)=0
We get rid of parentheses
-4F^2+F+50F-150=0
We add all the numbers together, and all the variables
-4F^2+51F-150=0
a = -4; b = 51; c = -150;
Δ = b2-4ac
Δ = 512-4·(-4)·(-150)
Δ = 201
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{-(51)-\sqrt{201}}{2*-4}=\frac{-51-\sqrt{201}}{-8} $
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+\sqrt{201}}{2*-4}=\frac{-51+\sqrt{201}}{-8} $

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