What are the solutions to the equation 12h^2-152h+325=0 ?

The solutions to the equation 12h^2-152h+325=0 are h=(38+sqrt(469))/6 ,h=(38-sqrt(469))/6

Find the zeros of 12h^2-152h+325=0

The zeros of 12h^2-152h+325=0 are h=(38+sqrt(469))/6 ,h=(38-sqrt(469))/6

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12 h^{2} - 152 h + 325 = 0

h = \frac{38 + 469}{6}, h = \frac{38 - 469}{6}

Decimal Notation

h = 9.94273 \dots, h = 2.72393 \dots

Solution steps

12 h^{2} - 152 h + 325 = 0

Solve with the quadratic formula

h_{1, 2} = \frac{- ( - 152 ) \pm ( - 152 ) ^{2} - 4 \cdot 12 \cdot 325}{2 \cdot 12}

(- 152)^{2} - 4 \cdot 12 \cdot 325 = 4469

h_{1, 2} = \frac{- ( - 152 ) \pm 4 469}{2 \cdot 12}

Separate the solutions

h_{1} = \frac{- ( - 152 ) + 4 469}{2 \cdot 12}, h_{2} = \frac{- ( - 152 ) - 4 469}{2 \cdot 12}

h = \frac{- ( - 152 ) + 4 469}{2 \cdot 12} : \frac{38 + 469}{6}

h = \frac{- ( - 152 ) - 4 469}{2 \cdot 12} : \frac{38 - 469}{6}

The solutions to the quadratic equation are:

h = \frac{38 + 469}{6}, h = \frac{38 - 469}{6}

- 16 t^{2} + 560 t = 0

16 x^{2} - 120 x + 180 = 0

x^{2} - 15 x + 26 = - 3 x - 6

8 x^{2} + 24 x - 20 = 0

2 x^{2} - 5 x - 7 = 0

What are the solutions to the equation 12h^2-152h+325=0 ?
The solutions to the equation 12h^2-152h+325=0 are h=(38+sqrt(469))/6 ,h=(38-sqrt(469))/6
Find the zeros of 12h^2-152h+325=0
The zeros of 12h^2-152h+325=0 are h=(38+sqrt(469))/6 ,h=(38-sqrt(469))/6