What are the solutions to the equation 2x^2-24x+33=0 ?

The solutions to the equation 2x^2-24x+33=0 are x=(12+sqrt(78))/2 ,x=(12-sqrt(78))/2

Find the zeros of 2x^2-24x+33=0

The zeros of 2x^2-24x+33=0 are x=(12+sqrt(78))/2 ,x=(12-sqrt(78))/2

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2 x^{2} - 24 x + 33 = 0

x = \frac{12 + 78}{2}, x = \frac{12 - 78}{2}

Decimal Notation

x = 10.41588 \dots, x = 1.58411 \dots

Solution steps

2 x^{2} - 24 x + 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 2 \cdot 33}{2 \cdot 2}

(- 24)^{2} - 4 \cdot 2 \cdot 33 = 278

x_{1, 2} = \frac{- ( - 24 ) \pm 2 78}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 2 78}{2 \cdot 2}, x_{2} = \frac{- ( - 24 ) - 2 78}{2 \cdot 2}

x = \frac{- ( - 24 ) + 2 78}{2 \cdot 2} : \frac{12 + 78}{2}

x = \frac{- ( - 24 ) - 2 78}{2 \cdot 2} : \frac{12 - 78}{2}

The solutions to the quadratic equation are:

x = \frac{12 + 78}{2}, x = \frac{12 - 78}{2}

5 y^{2} + 35 + 2 = 0

3 s^{2} - 9 s - 1 = 0

2 x^{2} = - 7 x - 6

h^{2} - 10 h - 4 = 0

w^{2} + 12 w

What are the solutions to the equation 2x^2-24x+33=0 ?
The solutions to the equation 2x^2-24x+33=0 are x=(12+sqrt(78))/2 ,x=(12-sqrt(78))/2
Find the zeros of 2x^2-24x+33=0
The zeros of 2x^2-24x+33=0 are x=(12+sqrt(78))/2 ,x=(12-sqrt(78))/2