What are the solutions to the equation d^2=4x^2-12x+9 ?

The solutions to the equation d^2=4x^2-12x+9 are x=(3+d)/2 ,x=(3-d)/2

Find the zeros of d^2=4x^2-12x+9

The zeros of d^2=4x^2-12x+9 are x=(3+d)/2 ,x=(3-d)/2

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d^{2} = 4 x^{2} - 12 x + 9

x = \frac{3 + d}{2}, x = \frac{3 - d}{2}

Solution steps

d^{2} = 4 x^{2} - 12 x + 9

Switch sides

4 x^{2} - 12 x + 9 = d^{2}

Move

d^{2}

to the left side

4 x^{2} - 12 x + 9 - d^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 ( 9 - d ^{2} )}{2 \cdot 4}

Simplify

(- 12)^{2} - 4 \cdot 4 (9 - d^{2}) : 4 d

x_{1, 2} = \frac{- ( - 12 ) \pm 4 d}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 d}{2 \cdot 4}, x_{2} = \frac{- ( - 12 ) - 4 d}{2 \cdot 4}

x = \frac{- ( - 12 ) + 4 d}{2 \cdot 4} : \frac{3 + d}{2}

x = \frac{- ( - 12 ) - 4 d}{2 \cdot 4} : \frac{3 - d}{2}

The solutions to the quadratic equation are:

x = \frac{3 + d}{2}, x = \frac{3 - d}{2}

0.2 x^{2} + 1.2 x - 32 = 0

x, y^{2} e^{1} + x^{2} = e (x^{2} - 1) - 1 e (x^{2} + 1)

9 x^{2} - 25 = d

x, 12 x^{2} - 15 x y = 0

x, 4 x^{2} + 2 y^{2} = 4 x y + 2 x y - 1

What are the solutions to the equation d^2=4x^2-12x+9 ?
The solutions to the equation d^2=4x^2-12x+9 are x=(3+d)/2 ,x=(3-d)/2
Find the zeros of d^2=4x^2-12x+9
The zeros of d^2=4x^2-12x+9 are x=(3+d)/2 ,x=(3-d)/2