What are the solutions to the equation 2x^2-12x+15=0 ?

The solutions to the equation 2x^2-12x+15=0 are x=(6+sqrt(6))/2 ,x=(6-sqrt(6))/2

Find the zeros of 2x^2-12x+15=0

The zeros of 2x^2-12x+15=0 are x=(6+sqrt(6))/2 ,x=(6-sqrt(6))/2

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2 x^{2} - 12 x + 15 = 0

x = \frac{6 + 6}{2}, x = \frac{6 - 6}{2}

Decimal Notation

x = 4.22474 \dots, x = 1.77525 \dots

Solution steps

2 x^{2} - 12 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 \cdot 15}{2 \cdot 2}

(- 12)^{2} - 4 \cdot 2 \cdot 15 = 26

x_{1, 2} = \frac{- ( - 12 ) \pm 2 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 6}{2 \cdot 2}, x_{2} = \frac{- ( - 12 ) - 2 6}{2 \cdot 2}

x = \frac{- ( - 12 ) + 2 6}{2 \cdot 2} : \frac{6 + 6}{2}

x = \frac{- ( - 12 ) - 2 6}{2 \cdot 2} : \frac{6 - 6}{2}

The solutions to the quadratic equation are:

x = \frac{6 + 6}{2}, x = \frac{6 - 6}{2}

1 - (\frac{7}{25})^{2}

(2 x + 3) + (x - 1) < 0

\frac{8 x ^{- 3} y}{24 y ^{- 4} x ^{2}}

- 4 + x \leq 2 (x - 5)

x^{4} + 3 x^{3} - 4 x^{2} - 12 x

What are the solutions to the equation 2x^2-12x+15=0 ?
The solutions to the equation 2x^2-12x+15=0 are x=(6+sqrt(6))/2 ,x=(6-sqrt(6))/2
Find the zeros of 2x^2-12x+15=0
The zeros of 2x^2-12x+15=0 are x=(6+sqrt(6))/2 ,x=(6-sqrt(6))/2