What are the solutions to the equation 20(t^2+16)=320t ?

The solutions to the equation 20(t^2+16)=320t are t=4(2+sqrt(3)),t=4(2-sqrt(3))

Find the zeros of 20(t^2+16)=320t

The zeros of 20(t^2+16)=320t are t=4(2+sqrt(3)),t=4(2-sqrt(3))

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20 (t^{2} + 16) = 320 t

t = 4 (2 + 3), t = 4 (2 - 3)

Decimal Notation

t = 14.92820 \dots, t = 1.07179 \dots

Solution steps

20 (t^{2} + 16) = 320 t

Expand

20 (t^{2} + 16) : 20 t^{2} + 320

20 t^{2} + 320 = 320 t

Move

320 t

to the left side

20 t^{2} + 320 - 320 t = 0

Write in the standard form

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

20 t^{2} - 320 t + 320 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 320 ) \pm ( - 320 ) ^{2} - 4 \cdot 20 \cdot 320}{2 \cdot 20}

(- 320)^{2} - 4 \cdot 20 \cdot 320 = 1603

t_{1, 2} = \frac{- ( - 320 ) \pm 160 3}{2 \cdot 20}

Separate the solutions

t_{1} = \frac{- ( - 320 ) + 160 3}{2 \cdot 20}, t_{2} = \frac{- ( - 320 ) - 160 3}{2 \cdot 20}

t = \frac{- ( - 320 ) + 160 3}{2 \cdot 20} : 4 (2 + 3)

t = \frac{- ( - 320 ) - 160 3}{2 \cdot 20} : 4 (2 - 3)

The solutions to the quadratic equation are:

t = 4 (2 + 3), t = 4 (2 - 3)

(2 t - 3)^{2} - 8 (2 t - 3) + 7 = 0

8 = 3 t^{2} - 5 t + 1

2 r^{2} + 6 r = - 12 - 2 r

5 (x + 5)^{2} = 60

(x - 3)^{3} = x^{3} - 63

What are the solutions to the equation 20(t^2+16)=320t ?
The solutions to the equation 20(t^2+16)=320t are t=4(2+sqrt(3)),t=4(2-sqrt(3))
Find the zeros of 20(t^2+16)=320t
The zeros of 20(t^2+16)=320t are t=4(2+sqrt(3)),t=4(2-sqrt(3))