What are the solutions to the equation (3x^2-248x)/4+961=0 ?

The solutions to the equation (3x^2-248x)/4+961=0 are x=62,x= 62/3

Find the zeros of (3x^2-248x)/4+961=0

The zeros of (3x^2-248x)/4+961=0 are x=62,x= 62/3

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\frac{3 x ^{2} - 248 x}{4} + 961 = 0

x = 62, x = \frac{62}{3}

Decimal Notation

x = 62, x = 20.66666 \dots

Solution steps

\frac{3 x ^{2} - 248 x}{4} + 961 = 0

Multiply both sides by

4

3 x^{2} - 248 x + 3844 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 248 ) \pm ( - 248 ) ^{2} - 4 \cdot 3 \cdot 3844}{2 \cdot 3}

(- 248)^{2} - 4 \cdot 3 \cdot 3844 = 124

x_{1, 2} = \frac{- ( - 248 ) \pm 124}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 248 ) + 124}{2 \cdot 3}, x_{2} = \frac{- ( - 248 ) - 124}{2 \cdot 3}

x = \frac{- ( - 248 ) + 124}{2 \cdot 3} : 62

x = \frac{- ( - 248 ) - 124}{2 \cdot 3} : \frac{62}{3}

The solutions to the quadratic equation are:

x = 62, x = \frac{62}{3}

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0 = - 3 x^{2} - 4 x + 5

\frac{π x ^{2} - 22 x ^{2} + 720 x}{9} = 0

7 x^{2} - 70 x + 175 = 0

What are the solutions to the equation (3x^2-248x)/4+961=0 ?
The solutions to the equation (3x^2-248x)/4+961=0 are x=62,x= 62/3
Find the zeros of (3x^2-248x)/4+961=0
The zeros of (3x^2-248x)/4+961=0 are x=62,x= 62/3