What are the solutions to the equation 2x^2-48x+288=0 ?

The solutions to the equation 2x^2-48x+288=0 are x=12

Find the zeros of 2x^2-48x+288=0

The zeros of 2x^2-48x+288=0 are x=12

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2 x^{2} - 48 x + 288 = 0

x = 12

Solution steps

2 x^{2} - 48 x + 288 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 2 \cdot 288}{2 \cdot 2}

(- 48)^{2} - 4 \cdot 2 \cdot 288 = 0

x_{1, 2} = \frac{- ( - 48 ) \pm 0}{2 \cdot 2}

x = \frac{- ( - 48 )}{2 \cdot 2}

\frac{- ( - 48 )}{2 \cdot 2} = 12

x = 12

The solution to the quadratic equation is:

x = 12

c^{2} + 3 c + 5 = 0

x^{2} - 6 = - 7 x

(3 x - 2)^{2} + 8 (3 x - 2) + 12 = 0

r^{2} + r + 0 = 0

x^{2} + 6 x - 247 = 0

What are the solutions to the equation 2x^2-48x+288=0 ?
The solutions to the equation 2x^2-48x+288=0 are x=12
Find the zeros of 2x^2-48x+288=0
The zeros of 2x^2-48x+288=0 are x=12