What are the solutions to the equation (4t-6)^2-12(4t-6)+20=0 ?

The solutions to the equation (4t-6)^2-12(4t-6)+20=0 are t=4,t=2

Find the zeros of (4t-6)^2-12(4t-6)+20=0

The zeros of (4t-6)^2-12(4t-6)+20=0 are t=4,t=2

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(4 t - 6)^{2} - 12 (4 t - 6) + 20 = 0

t = 4, t = 2

Solution steps

(4 t - 6)^{2} - 12 (4 t - 6) + 20 = 0

Expand

(4 t - 6)^{2} - 12 (4 t - 6) + 20 : 16 t^{2} - 96 t + 128

16 t^{2} - 96 t + 128 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 96 ) \pm ( - 96 ) ^{2} - 4 \cdot 16 \cdot 128}{2 \cdot 16}

(- 96)^{2} - 4 \cdot 16 \cdot 128 = 32

t_{1, 2} = \frac{- ( - 96 ) \pm 32}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 96 ) + 32}{2 \cdot 16}, t_{2} = \frac{- ( - 96 ) - 32}{2 \cdot 16}

t = \frac{- ( - 96 ) + 32}{2 \cdot 16} : 4

t = \frac{- ( - 96 ) - 32}{2 \cdot 16} : 2

The solutions to the quadratic equation are:

t = 4, t = 2

(\frac{1}{2} + \frac{1}{3})^{2}

\frac{- x + 5}{x - 8} \geq 0

x^{3} - 6 x^{2} - x + 6 \geq 0

x^{2} - 18 x + 78 = 0

(4 x)^{- \frac{1}{2}} \cdot (9 x)^{\frac{1}{2}}

What are the solutions to the equation (4t-6)^2-12(4t-6)+20=0 ?
The solutions to the equation (4t-6)^2-12(4t-6)+20=0 are t=4,t=2
Find the zeros of (4t-6)^2-12(4t-6)+20=0
The zeros of (4t-6)^2-12(4t-6)+20=0 are t=4,t=2