What are the solutions to the equation -4t^2-20t+3=0 ?

The solutions to the equation -4t^2-20t+3=0 are t=-(5+2sqrt(7))/2 ,t=(2sqrt(7)-5)/2

Find the zeros of -4t^2-20t+3=0

The zeros of -4t^2-20t+3=0 are t=-(5+2sqrt(7))/2 ,t=(2sqrt(7)-5)/2

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- 4 t^{2} - 20 t + 3 = 0

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

Decimal Notation

t = - 5.14575 \dots, t = 0.14575 \dots

Solution steps

- 4 t^{2} - 20 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 4 ) \cdot 3}{2 ( - 4 )}

(- 20)^{2} - 4 (- 4) \cdot 3 = 87

t_{1, 2} = \frac{- ( - 20 ) \pm 8 7}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- ( - 20 ) + 8 7}{2 ( - 4 )}, t_{2} = \frac{- ( - 20 ) - 8 7}{2 ( - 4 )}

t = \frac{- ( - 20 ) + 8 7}{2 ( - 4 )} : - \frac{5 + 2 7}{2}

t = \frac{- ( - 20 ) - 8 7}{2 ( - 4 )} : \frac{2 7 - 5}{2}

The solutions to the quadratic equation are:

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

7 x^{2} - 8 x - 12 = 0

3 n^{2} - 20 n + 12 = 0

4 x^{2} + 9 + 12 x = 0

x^{2} + 4 x + 11 = 0

(17 x - 1)^{2} - 4 (17 x - 1) - 5 = 0

What are the solutions to the equation -4t^2-20t+3=0 ?
The solutions to the equation -4t^2-20t+3=0 are t=-(5+2sqrt(7))/2 ,t=(2sqrt(7)-5)/2
Find the zeros of -4t^2-20t+3=0
The zeros of -4t^2-20t+3=0 are t=-(5+2sqrt(7))/2 ,t=(2sqrt(7)-5)/2