What are the solutions to the equation 2x^2-24x+52=0 ?

The solutions to the equation 2x^2-24x+52=0 are x=6+sqrt(10),x=6-sqrt(10)

Find the zeros of 2x^2-24x+52=0

The zeros of 2x^2-24x+52=0 are x=6+sqrt(10),x=6-sqrt(10)

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2 x^{2} - 24 x + 52 = 0

x = 6 + 10, x = 6 - 10

Decimal Notation

x = 9.16227 \dots, x = 2.83772 \dots

Solution steps

2 x^{2} - 24 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 2 \cdot 52}{2 \cdot 2}

(- 24)^{2} - 4 \cdot 2 \cdot 52 = 410

x_{1, 2} = \frac{- ( - 24 ) \pm 4 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 4 10}{2 \cdot 2}, x_{2} = \frac{- ( - 24 ) - 4 10}{2 \cdot 2}

x = \frac{- ( - 24 ) + 4 10}{2 \cdot 2} : 6 + 10

x = \frac{- ( - 24 ) - 4 10}{2 \cdot 2} : 6 - 10

The solutions to the quadratic equation are:

x = 6 + 10, x = 6 - 10

- 3 - 0.5 x + 13 = 0.3 x + 6.2

16 - x^{2} = - 19

4 x^{2} + 8 x - 45 = 0

12 x^{2} - 12

x^{2} = 0.34

What are the solutions to the equation 2x^2-24x+52=0 ?
The solutions to the equation 2x^2-24x+52=0 are x=6+sqrt(10),x=6-sqrt(10)
Find the zeros of 2x^2-24x+52=0
The zeros of 2x^2-24x+52=0 are x=6+sqrt(10),x=6-sqrt(10)