What are the solutions to the equation 2x^2-8x+16=38 ?

The solutions to the equation 2x^2-8x+16=38 are x=2+sqrt(15),x=2-sqrt(15)

Find the zeros of 2x^2-8x+16=38

The zeros of 2x^2-8x+16=38 are x=2+sqrt(15),x=2-sqrt(15)

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2 x^{2} - 8 x + 16 = 38

x = 2 + 15, x = 2 - 15

Decimal Notation

x = 5.87298 \dots, x = - 1.87298 \dots

Solution steps

2 x^{2} - 8 x + 16 = 38

Move

38

to the left side

2 x^{2} - 8 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 22 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 22) = 415

x_{1, 2} = \frac{- ( - 8 ) \pm 4 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 15}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 4 15}{2 \cdot 2}

x = \frac{- ( - 8 ) + 4 15}{2 \cdot 2} : 2 + 15

x = \frac{- ( - 8 ) - 4 15}{2 \cdot 2} : 2 - 15

The solutions to the quadratic equation are:

x = 2 + 15, x = 2 - 15

x^{2} - 21 x + 104 = 0

y, x = \frac{1}{2} y^{2} + \frac{1}{4}

x^{2} + 3 x + 3 = - x^{2} + x + 7

(x - 9) (x + 5) = 0

x - \frac{x ^{2}}{4} = 0

What are the solutions to the equation 2x^2-8x+16=38 ?
The solutions to the equation 2x^2-8x+16=38 are x=2+sqrt(15),x=2-sqrt(15)
Find the zeros of 2x^2-8x+16=38
The zeros of 2x^2-8x+16=38 are x=2+sqrt(15),x=2-sqrt(15)