What are the solutions to the equation 2t^2-20t+48=0 ?

The solutions to the equation 2t^2-20t+48=0 are t=6,t=4

Find the zeros of 2t^2-20t+48=0

The zeros of 2t^2-20t+48=0 are t=6,t=4

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2 t^{2} - 20 t + 48 = 0

t = 6, t = 4

Solution steps

2 t^{2} - 20 t + 48 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 48}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 48 = 4

t_{1, 2} = \frac{- ( - 20 ) \pm 4}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 20 ) + 4}{2 \cdot 2}, t_{2} = \frac{- ( - 20 ) - 4}{2 \cdot 2}

t = \frac{- ( - 20 ) + 4}{2 \cdot 2} : 6

t = \frac{- ( - 20 ) - 4}{2 \cdot 2} : 4

The solutions to the quadratic equation are:

t = 6, t = 4

x^{2} - 6 x + 8 = x - 4

4 (6) + 7 (2) = - 19

4^{2} + 6^{2} = c^{2}

\frac{y}{5} - 3 = \frac{y}{2}

What are the solutions to the equation 2t^2-20t+48=0 ?
The solutions to the equation 2t^2-20t+48=0 are t=6,t=4
Find the zeros of 2t^2-20t+48=0
The zeros of 2t^2-20t+48=0 are t=6,t=4