What are the solutions to the equation 5t^2-40t=-15 ?

The solutions to the equation 5t^2-40t=-15 are t=4+sqrt(13),t=4-sqrt(13)

Find the zeros of 5t^2-40t=-15

The zeros of 5t^2-40t=-15 are t=4+sqrt(13),t=4-sqrt(13)

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5 t^{2} - 40 t = - 15

t = 4 + 13, t = 4 - 13

Decimal Notation

t = 7.60555 \dots, t = 0.39444 \dots

Solution steps

5 t^{2} - 40 t = - 15

Move

15

to the left side

5 t^{2} - 40 t + 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 5 \cdot 15}{2 \cdot 5}

(- 40)^{2} - 4 \cdot 5 \cdot 15 = 1013

t_{1, 2} = \frac{- ( - 40 ) \pm 10 13}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 40 ) + 10 13}{2 \cdot 5}, t_{2} = \frac{- ( - 40 ) - 10 13}{2 \cdot 5}

t = \frac{- ( - 40 ) + 10 13}{2 \cdot 5} : 4 + 13

t = \frac{- ( - 40 ) - 10 13}{2 \cdot 5} : 4 - 13

The solutions to the quadratic equation are:

t = 4 + 13, t = 4 - 13

2 k^{2} - 16 k^{2} - 3 k + 24

\frac{1}{4} (2)

1 - \frac{x}{2} > 3

- 12 (x + 6) = 4 (x - 6)

- 3 h + 28 = - 17 h

What are the solutions to the equation 5t^2-40t=-15 ?
The solutions to the equation 5t^2-40t=-15 are t=4+sqrt(13),t=4-sqrt(13)
Find the zeros of 5t^2-40t=-15
The zeros of 5t^2-40t=-15 are t=4+sqrt(13),t=4-sqrt(13)