What are the solutions to the equation 2x^2-16x+12=0 ?

The solutions to the equation 2x^2-16x+12=0 are x=4+sqrt(10),x=4-sqrt(10)

Find the zeros of 2x^2-16x+12=0

The zeros of 2x^2-16x+12=0 are x=4+sqrt(10),x=4-sqrt(10)

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2 x^{2} - 16 x + 12 = 0

x = 4 + 10, x = 4 - 10

Decimal Notation

x = 7.16227 \dots, x = 0.83772 \dots

Solution steps

2 x^{2} - 16 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 \cdot 12}{2 \cdot 2}

(- 16)^{2} - 4 \cdot 2 \cdot 12 = 410

x_{1, 2} = \frac{- ( - 16 ) \pm 4 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4 10}{2 \cdot 2}, x_{2} = \frac{- ( - 16 ) - 4 10}{2 \cdot 2}

x = \frac{- ( - 16 ) + 4 10}{2 \cdot 2} : 4 + 10

x = \frac{- ( - 16 ) - 4 10}{2 \cdot 2} : 4 - 10

The solutions to the quadratic equation are:

x = 4 + 10, x = 4 - 10

4 b^{2} + 12 b - 3 = 5 b

x, x^{2} = y^{2} + z^{2}

16 t^{2} + 8 t - 8 = - 5

- 3 v^{2} + 10 v + 5 = - 8 v^{2}

15 d^{2} - 29 d + 8 = - 4

What are the solutions to the equation 2x^2-16x+12=0 ?
The solutions to the equation 2x^2-16x+12=0 are x=4+sqrt(10),x=4-sqrt(10)
Find the zeros of 2x^2-16x+12=0
The zeros of 2x^2-16x+12=0 are x=4+sqrt(10),x=4-sqrt(10)