H(x)=(-2x+3)(-x+3)

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Solution for H(x)=(-2x+3)(-x+3) equation:



(H)=(-2H+3)(-H+3)
We move all terms to the left:
(H)-((-2H+3)(-H+3))=0
We add all the numbers together, and all the variables
H-((-2H+3)(-1H+3))=0
We multiply parentheses ..
-((+2H^2-6H-3H+9))+H=0
We calculate terms in parentheses: -((+2H^2-6H-3H+9)), so:
(+2H^2-6H-3H+9)
We get rid of parentheses
2H^2-6H-3H+9
We add all the numbers together, and all the variables
2H^2-9H+9
Back to the equation:
-(2H^2-9H+9)
We add all the numbers together, and all the variables
H-(2H^2-9H+9)=0
We get rid of parentheses
-2H^2+H+9H-9=0
We add all the numbers together, and all the variables
-2H^2+10H-9=0
a = -2; b = 10; c = -9;
Δ = b2-4ac
Δ = 102-4·(-2)·(-9)
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{7}}{2*-2}=\frac{-10-2\sqrt{7}}{-4} $
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{7}}{2*-2}=\frac{-10+2\sqrt{7}}{-4} $

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