What are the solutions to the equation (5t-9)^2-7(5t-9)+6=0 ?

The solutions to the equation (5t-9)^2-7(5t-9)+6=0 are t=3,t=2

Find the zeros of (5t-9)^2-7(5t-9)+6=0

The zeros of (5t-9)^2-7(5t-9)+6=0 are t=3,t=2

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(5 t - 9)^{2} - 7 (5 t - 9) + 6 = 0

t = 3, t = 2

Solution steps

(5 t - 9)^{2} - 7 (5 t - 9) + 6 = 0

Expand

(5 t - 9)^{2} - 7 (5 t - 9) + 6 : 25 t^{2} - 125 t + 150

25 t^{2} - 125 t + 150 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 125 ) \pm ( - 125 ) ^{2} - 4 \cdot 25 \cdot 150}{2 \cdot 25}

(- 125)^{2} - 4 \cdot 25 \cdot 150 = 25

t_{1, 2} = \frac{- ( - 125 ) \pm 25}{2 \cdot 25}

Separate the solutions

t_{1} = \frac{- ( - 125 ) + 25}{2 \cdot 25}, t_{2} = \frac{- ( - 125 ) - 25}{2 \cdot 25}

t = \frac{- ( - 125 ) + 25}{2 \cdot 25} : 3

t = \frac{- ( - 125 ) - 25}{2 \cdot 25} : 2

The solutions to the quadratic equation are:

t = 3, t = 2

x (10 - x) = 40

x, 2 x y = x^{2}

x^{2} + 17 x - 30 = 0

0.25 x^{2} - 2 x + 3 = 0

x^{2} + 12 x - 15 = 0

What are the solutions to the equation (5t-9)^2-7(5t-9)+6=0 ?
The solutions to the equation (5t-9)^2-7(5t-9)+6=0 are t=3,t=2
Find the zeros of (5t-9)^2-7(5t-9)+6=0
The zeros of (5t-9)^2-7(5t-9)+6=0 are t=3,t=2