What are the solutions to the equation 3X^2+40X-180=0 ?

The solutions to the equation 3X^2+40X-180=0 are X=(2(sqrt(235)-10))/3 ,X=-(2(10+sqrt(235)))/3

Find the zeros of 3X^2+40X-180=0

The zeros of 3X^2+40X-180=0 are X=(2(sqrt(235)-10))/3 ,X=-(2(10+sqrt(235)))/3

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3 X^{2} + 40 X - 180 = 0

X = \frac{2 ( 235 - 10 )}{3}, X = - \frac{2 ( 10 + 235 )}{3}

Decimal Notation

X = 3.55313 \dots, X = - 16.88647 \dots

Solution steps

3 X^{2} + 40 X - 180 = 0

Solve with the quadratic formula

X_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 3 ( - 180 )}{2 \cdot 3}

4 0^{2} - 4 \cdot 3 (- 180) = 4235

X_{1, 2} = \frac{- 40 \pm 4 235}{2 \cdot 3}

Separate the solutions

X_{1} = \frac{- 40 + 4 235}{2 \cdot 3}, X_{2} = \frac{- 40 - 4 235}{2 \cdot 3}

X = \frac{- 40 + 4 235}{2 \cdot 3} : \frac{2 ( 235 - 10 )}{3}

X = \frac{- 40 - 4 235}{2 \cdot 3} : - \frac{2 ( 10 + 235 )}{3}

The solutions to the quadratic equation are:

X = \frac{2 ( 235 - 10 )}{3}, X = - \frac{2 ( 10 + 235 )}{3}

x^{2} + 4 x - 11 = 1

x^{2} + 127 = 23 x + 7

x^{2} - 2 x - 21 = 0

3 t^{2} + 6 t = 6

7 = 0.003 x^{2} + 0.001 x

What are the solutions to the equation 3X^2+40X-180=0 ?
The solutions to the equation 3X^2+40X-180=0 are X=(2(sqrt(235)-10))/3 ,X=-(2(10+sqrt(235)))/3
Find the zeros of 3X^2+40X-180=0
The zeros of 3X^2+40X-180=0 are X=(2(sqrt(235)-10))/3 ,X=-(2(10+sqrt(235)))/3