What are the solutions to the equation 4t^2=20t+150 ?

The solutions to the equation 4t^2=20t+150 are t=(5(1+sqrt(7)))/2 ,t=(5(1-sqrt(7)))/2

Find the zeros of 4t^2=20t+150

The zeros of 4t^2=20t+150 are t=(5(1+sqrt(7)))/2 ,t=(5(1-sqrt(7)))/2

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4 t^{2} = 20 t + 150

t = \frac{5 ( 1 + 7 )}{2}, t = \frac{5 ( 1 - 7 )}{2}

Decimal Notation

t = 9.11437 \dots, t = - 4.11437 \dots

Solution steps

4 t^{2} = 20 t + 150

Move

150

to the left side

4 t^{2} - 150 = 20 t

Move

20 t

to the left side

4 t^{2} - 150 - 20 t = 0

Write in the standard form

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

4 t^{2} - 20 t - 150 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 ( - 150 )}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 (- 150) = 207

t_{1, 2} = \frac{- ( - 20 ) \pm 20 7}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- ( - 20 ) + 20 7}{2 \cdot 4}, t_{2} = \frac{- ( - 20 ) - 20 7}{2 \cdot 4}

t = \frac{- ( - 20 ) + 20 7}{2 \cdot 4} : \frac{5 ( 1 + 7 )}{2}

t = \frac{- ( - 20 ) - 20 7}{2 \cdot 4} : \frac{5 ( 1 - 7 )}{2}

The solutions to the quadratic equation are:

t = \frac{5 ( 1 + 7 )}{2}, t = \frac{5 ( 1 - 7 )}{2}

x^{3} + x^{2} y^{4}

13 < m - 25

\frac{\frac{4}{5} - \frac{5}{3}}{\frac{3}{4} - \frac{5}{2}}

π 7^{2}

\frac{b ^{4}}{b ^{2}}

What are the solutions to the equation 4t^2=20t+150 ?
The solutions to the equation 4t^2=20t+150 are t=(5(1+sqrt(7)))/2 ,t=(5(1-sqrt(7)))/2
Find the zeros of 4t^2=20t+150
The zeros of 4t^2=20t+150 are t=(5(1+sqrt(7)))/2 ,t=(5(1-sqrt(7)))/2