What are the solutions to the equation 1.2x^2-0.6x-10.2=0 ?

The solutions to the equation 1.2x^2-0.6x-10.2=0 are x=(1+sqrt(137))/4 ,x=(1-sqrt(137))/4

Find the zeros of 1.2x^2-0.6x-10.2=0

The zeros of 1.2x^2-0.6x-10.2=0 are x=(1+sqrt(137))/4 ,x=(1-sqrt(137))/4

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1.2 x^{2} - 0.6 x - 10.2 = 0

x = \frac{1 + 137}{4}, x = \frac{1 - 137}{4}

Decimal

x = 3.17617 \dots, x = - 2.67617 \dots

Solution steps

1.2 x^{2} - 0.6 x - 10.2 = 0

Multiply both sides by

10

12 x^{2} - 6 x - 102 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 12 ( - 102 )}{2 \cdot 12}

(- 6)^{2} - 4 \cdot 12 (- 102) = 6137

x_{1, 2} = \frac{- ( - 6 ) \pm 6 137}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6 137}{2 \cdot 12}, x_{2} = \frac{- ( - 6 ) - 6 137}{2 \cdot 12}

x = \frac{- ( - 6 ) + 6 137}{2 \cdot 12} : \frac{1 + 137}{4}

x = \frac{- ( - 6 ) - 6 137}{2 \cdot 12} : \frac{1 - 137}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 137}{4}, x = \frac{1 - 137}{4}

64 - 16 x = 32 x^{2}

x^{2} + 8 x - 240 = 0

- 2 w^{2} + 6 w = 5

4 x^{2} = x^{2}

(\frac{1}{3})^{2} + x^{2} = 1

What are the solutions to the equation 1.2x^2-0.6x-10.2=0 ?
The solutions to the equation 1.2x^2-0.6x-10.2=0 are x=(1+sqrt(137))/4 ,x=(1-sqrt(137))/4
Find the zeros of 1.2x^2-0.6x-10.2=0
The zeros of 1.2x^2-0.6x-10.2=0 are x=(1+sqrt(137))/4 ,x=(1-sqrt(137))/4