What are the solutions to the equation 16t^2-235t+151=0 ?

The solutions to the equation 16t^2-235t+151=0 are t=(235+sqrt(45561))/(32),t=(235-sqrt(45561))/(32)

Find the zeros of 16t^2-235t+151=0

The zeros of 16t^2-235t+151=0 are t=(235+sqrt(45561))/(32),t=(235-sqrt(45561))/(32)

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16 t^{2} - 235 t + 151 = 0

t = \frac{235 + 45561}{32}, t = \frac{235 - 45561}{32}

Decimal Notation

t = 14.01406 \dots, t = 0.67343 \dots

Solution steps

16 t^{2} - 235 t + 151 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 235 ) \pm ( - 235 ) ^{2} - 4 \cdot 16 \cdot 151}{2 \cdot 16}

(- 235)^{2} - 4 \cdot 16 \cdot 151 = 45561

t_{1, 2} = \frac{- ( - 235 ) \pm 45561}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 235 ) + 45561}{2 \cdot 16}, t_{2} = \frac{- ( - 235 ) - 45561}{2 \cdot 16}

t = \frac{- ( - 235 ) + 45561}{2 \cdot 16} : \frac{235 + 45561}{32}

t = \frac{- ( - 235 ) - 45561}{2 \cdot 16} : \frac{235 - 45561}{32}

The solutions to the quadratic equation are:

t = \frac{235 + 45561}{32}, t = \frac{235 - 45561}{32}

- 4 x + 35 > 2

8 - 5 ∣ 2 x + 10 ∣ = - 32

x^{2} + 3 = 39

(- 9)^{\frac{1}{2}}

9 (x + 1) (x - 1) + 7 (x + 1)^{2} = (4 x - 3)^{2} - 30

What are the solutions to the equation 16t^2-235t+151=0 ?
The solutions to the equation 16t^2-235t+151=0 are t=(235+sqrt(45561))/(32),t=(235-sqrt(45561))/(32)
Find the zeros of 16t^2-235t+151=0
The zeros of 16t^2-235t+151=0 are t=(235+sqrt(45561))/(32),t=(235-sqrt(45561))/(32)