What are the solutions to the equation 150=-6x^2+100x-180 ?

The solutions to the equation 150=-6x^2+100x-180 are x=(25-sqrt(130))/3 ,x=(25+sqrt(130))/3

Find the zeros of 150=-6x^2+100x-180

The zeros of 150=-6x^2+100x-180 are x=(25-sqrt(130))/3 ,x=(25+sqrt(130))/3

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150 = - 6 x^{2} + 100 x - 180

x = \frac{25 - 130}{3}, x = \frac{25 + 130}{3}

Decimal Notation

x = 4.53274 \dots, x = 12.13391 \dots

Solution steps

150 = - 6 x^{2} + 100 x - 180

Switch sides

- 6 x^{2} + 100 x - 180 = 150

Move

150

to the left side

- 6 x^{2} + 100 x - 330 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 ( - 6 ) ( - 330 )}{2 ( - 6 )}

10 0^{2} - 4 (- 6) (- 330) = 4130

x_{1, 2} = \frac{- 100 \pm 4 130}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- 100 + 4 130}{2 ( - 6 )}, x_{2} = \frac{- 100 - 4 130}{2 ( - 6 )}

x = \frac{- 100 + 4 130}{2 ( - 6 )} : \frac{25 - 130}{3}

x = \frac{- 100 - 4 130}{2 ( - 6 )} : \frac{25 + 130}{3}

The solutions to the quadratic equation are:

x = \frac{25 - 130}{3}, x = \frac{25 + 130}{3}

(D + 5) 3 D = - \frac{2}{4} (6 - 20 D) + 3

2 m^{2} + m = 0

x^{2} = 2 x + 13

x^{2} = 2 x + 25

x^{2} + 9 x + 7 = 0

What are the solutions to the equation 150=-6x^2+100x-180 ?
The solutions to the equation 150=-6x^2+100x-180 are x=(25-sqrt(130))/3 ,x=(25+sqrt(130))/3
Find the zeros of 150=-6x^2+100x-180
The zeros of 150=-6x^2+100x-180 are x=(25-sqrt(130))/3 ,x=(25+sqrt(130))/3