What are the solutions to the equation 2.7t^2-40t+7.5=0 ?

The solutions to the equation 2.7t^2-40t+7.5=0 are t=(5(40+7sqrt(31)))/(27),t=(5(40-7sqrt(31)))/(27)

Find the zeros of 2.7t^2-40t+7.5=0

The zeros of 2.7t^2-40t+7.5=0 are t=(5(40+7sqrt(31)))/(27),t=(5(40-7sqrt(31)))/(27)

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2.7 t^{2} - 40 t + 7.5 = 0

t = \frac{5 ( 40 + 7 31 )}{27}, t = \frac{5 ( 40 - 7 31 )}{27}

Decimal Notation

t = 14.62487 \dots, t = 0.18993 \dots

Solution steps

2.7 t^{2} - 40 t + 7.5 = 0

Multiply both sides by

10

27 t^{2} - 400 t + 75 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 400 ) \pm ( - 400 ) ^{2} - 4 \cdot 27 \cdot 75}{2 \cdot 27}

(- 400)^{2} - 4 \cdot 27 \cdot 75 = 7031

t_{1, 2} = \frac{- ( - 400 ) \pm 70 31}{2 \cdot 27}

Separate the solutions

t_{1} = \frac{- ( - 400 ) + 70 31}{2 \cdot 27}, t_{2} = \frac{- ( - 400 ) - 70 31}{2 \cdot 27}

t = \frac{- ( - 400 ) + 70 31}{2 \cdot 27} : \frac{5 ( 40 + 7 31 )}{27}

t = \frac{- ( - 400 ) - 70 31}{2 \cdot 27} : \frac{5 ( 40 - 7 31 )}{27}

The solutions to the quadratic equation are:

t = \frac{5 ( 40 + 7 31 )}{27}, t = \frac{5 ( 40 - 7 31 )}{27}

K^{2} - 25 h^{2}

\frac{2}{5} + \frac{3}{5}

x + \frac{7}{8} = \frac{1}{4}

\frac{1}{2} (x - 4) = - 8

\frac{p}{- 36} < 6

What are the solutions to the equation 2.7t^2-40t+7.5=0 ?
The solutions to the equation 2.7t^2-40t+7.5=0 are t=(5(40+7sqrt(31)))/(27),t=(5(40-7sqrt(31)))/(27)
Find the zeros of 2.7t^2-40t+7.5=0
The zeros of 2.7t^2-40t+7.5=0 are t=(5(40+7sqrt(31)))/(27),t=(5(40-7sqrt(31)))/(27)