What are the solutions to the equation 4.9t^2-20t+15=0 ?

The solutions to the equation 4.9t^2-20t+15=0 are t=(5(20+sqrt(106)))/(49),t=(5(20-sqrt(106)))/(49)

Find the zeros of 4.9t^2-20t+15=0

The zeros of 4.9t^2-20t+15=0 are t=(5(20+sqrt(106)))/(49),t=(5(20-sqrt(106)))/(49)

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4.9 t^{2} - 20 t + 15 = 0

t = \frac{5 ( 20 + 106 )}{49}, t = \frac{5 ( 20 - 106 )}{49}

Decimal Notation

t = 3.09139 \dots, t = 0.99024 \dots

Solution steps

4.9 t^{2} - 20 t + 15 = 0

Multiply both sides by

10

49 t^{2} - 200 t + 150 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 200 ) \pm ( - 200 ) ^{2} - 4 \cdot 49 \cdot 150}{2 \cdot 49}

(- 200)^{2} - 4 \cdot 49 \cdot 150 = 10106

t_{1, 2} = \frac{- ( - 200 ) \pm 10 106}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- ( - 200 ) + 10 106}{2 \cdot 49}, t_{2} = \frac{- ( - 200 ) - 10 106}{2 \cdot 49}

t = \frac{- ( - 200 ) + 10 106}{2 \cdot 49} : \frac{5 ( 20 + 106 )}{49}

t = \frac{- ( - 200 ) - 10 106}{2 \cdot 49} : \frac{5 ( 20 - 106 )}{49}

The solutions to the quadratic equation are:

t = \frac{5 ( 20 + 106 )}{49}, t = \frac{5 ( 20 - 106 )}{49}

21 - 8 k - 2 k^{2} = 2 k^{2}

x^{2} + 12 x + 20 = 0

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

y, 9 x = y^{2} + 18

(1 - (- 1)) x^{2} - (- 1 + 1) x - 2 = 0

What are the solutions to the equation 4.9t^2-20t+15=0 ?
The solutions to the equation 4.9t^2-20t+15=0 are t=(5(20+sqrt(106)))/(49),t=(5(20-sqrt(106)))/(49)
Find the zeros of 4.9t^2-20t+15=0
The zeros of 4.9t^2-20t+15=0 are t=(5(20+sqrt(106)))/(49),t=(5(20-sqrt(106)))/(49)