What are the solutions to the equation (2t-3)^2-8(2t-3)+7=0 ?

The solutions to the equation (2t-3)^2-8(2t-3)+7=0 are t=5,t=2

Find the zeros of (2t-3)^2-8(2t-3)+7=0

The zeros of (2t-3)^2-8(2t-3)+7=0 are t=5,t=2

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(2 t - 3)^{2} - 8 (2 t - 3) + 7 = 0

t = 5, t = 2

Solution steps

(2 t - 3)^{2} - 8 (2 t - 3) + 7 = 0

Expand

(2 t - 3)^{2} - 8 (2 t - 3) + 7 : 4 t^{2} - 28 t + 40

4 t^{2} - 28 t + 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 4 \cdot 40}{2 \cdot 4}

(- 28)^{2} - 4 \cdot 4 \cdot 40 = 12

t_{1, 2} = \frac{- ( - 28 ) \pm 12}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- ( - 28 ) + 12}{2 \cdot 4}, t_{2} = \frac{- ( - 28 ) - 12}{2 \cdot 4}

t = \frac{- ( - 28 ) + 12}{2 \cdot 4} : 5

t = \frac{- ( - 28 ) - 12}{2 \cdot 4} : 2

The solutions to the quadratic equation are:

t = 5, t = 2

8 = 3 t^{2} - 5 t + 1

2 r^{2} + 6 r = - 12 - 2 r

5 (x + 5)^{2} = 60

(x - 3)^{3} = x^{3} - 63

x^{2} + 16 x = 7

What are the solutions to the equation (2t-3)^2-8(2t-3)+7=0 ?
The solutions to the equation (2t-3)^2-8(2t-3)+7=0 are t=5,t=2
Find the zeros of (2t-3)^2-8(2t-3)+7=0
The zeros of (2t-3)^2-8(2t-3)+7=0 are t=5,t=2