What are the solutions to the equation 10-8t+0.8t^2=0 ?

The solutions to the equation 10-8t+0.8t^2=0 are t=(5(2+sqrt(2)))/2 ,t=(5(2-sqrt(2)))/2

Find the zeros of 10-8t+0.8t^2=0

The zeros of 10-8t+0.8t^2=0 are t=(5(2+sqrt(2)))/2 ,t=(5(2-sqrt(2)))/2

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10 - 8 t + 0.8 t^{2} = 0

t = \frac{5 ( 2 + 2 )}{2}, t = \frac{5 ( 2 - 2 )}{2}

Decimal Notation

t = 8.53553 \dots, t = 1.46446 \dots

Solution steps

10 - 8 t + 0.8 t^{2} = 0

Multiply both sides by

10

100 - 80 t + 8 t^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

8 t^{2} - 80 t + 100 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 80 ) \pm ( - 80 ) ^{2} - 4 \cdot 8 \cdot 100}{2 \cdot 8}

(- 80)^{2} - 4 \cdot 8 \cdot 100 = 402

t_{1, 2} = \frac{- ( - 80 ) \pm 40 2}{2 \cdot 8}

Separate the solutions

t_{1} = \frac{- ( - 80 ) + 40 2}{2 \cdot 8}, t_{2} = \frac{- ( - 80 ) - 40 2}{2 \cdot 8}

t = \frac{- ( - 80 ) + 40 2}{2 \cdot 8} : \frac{5 ( 2 + 2 )}{2}

t = \frac{- ( - 80 ) - 40 2}{2 \cdot 8} : \frac{5 ( 2 - 2 )}{2}

The solutions to the quadratic equation are:

t = \frac{5 ( 2 + 2 )}{2}, t = \frac{5 ( 2 - 2 )}{2}

- 9 h > (- 63)

2 x^{4} + 12 x^{3} - 8 x^{2} - 48 x

0 < ∣ - x - 6 ∣ < \frac{1}{2}

t^{3} + t^{2} - 2 t

(2 x + 7)^{2} = 64

What are the solutions to the equation 10-8t+0.8t^2=0 ?
The solutions to the equation 10-8t+0.8t^2=0 are t=(5(2+sqrt(2)))/2 ,t=(5(2-sqrt(2)))/2
Find the zeros of 10-8t+0.8t^2=0
The zeros of 10-8t+0.8t^2=0 are t=(5(2+sqrt(2)))/2 ,t=(5(2-sqrt(2)))/2