What are the solutions to the equation (12t^2-10t+5)+(10t-15)=0 ?

The solutions to the equation (12t^2-10t+5)+(10t-15)=0 are t=sqrt(5/6),t=-sqrt(5/6)

Find the zeros of (12t^2-10t+5)+(10t-15)=0

The zeros of (12t^2-10t+5)+(10t-15)=0 are t=sqrt(5/6),t=-sqrt(5/6)

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(12 t^{2} - 10 t + 5) + (10 t - 15) = 0

t = \frac{5}{6}, t = - \frac{5}{6}

Decimal Notation

t = 0.91287 \dots, t = - 0.91287 \dots

Solution steps

(12 t^{2} - 10 t + 5) + (10 t - 15) = 0

Refine

12 t^{2} - 10 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 12 ( - 10 )}{2 \cdot 12}

0^{2} - 4 \cdot 12 (- 10) = 430

t_{1, 2} = \frac{- 0 \pm 4 30}{2 \cdot 12}

Separate the solutions

t_{1} = \frac{- 0 + 4 30}{2 \cdot 12}, t_{2} = \frac{- 0 - 4 30}{2 \cdot 12}

t = \frac{- 0 + 4 30}{2 \cdot 12} : \frac{5}{6}

t = \frac{- 0 - 4 30}{2 \cdot 12} : - \frac{5}{6}

The solutions to the quadratic equation are:

t = \frac{5}{6}, t = - \frac{5}{6}

x, x^{2} + x y = 20 x

x, x^{2} + x y = 20 y

x = y^{2}, y

x^{2} + x + 10 = 20

4 x^{2} - 28 x + 48 = 0

What are the solutions to the equation (12t^2-10t+5)+(10t-15)=0 ?
The solutions to the equation (12t^2-10t+5)+(10t-15)=0 are t=sqrt(5/6),t=-sqrt(5/6)
Find the zeros of (12t^2-10t+5)+(10t-15)=0
The zeros of (12t^2-10t+5)+(10t-15)=0 are t=sqrt(5/6),t=-sqrt(5/6)