What are the solutions to the equation 0=4x^2-33x+63 ?

The solutions to the equation 0=4x^2-33x+63 are x= 21/4 ,x=3

Find the zeros of 0=4x^2-33x+63

The zeros of 0=4x^2-33x+63 are x= 21/4 ,x=3

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0 = 4 x^{2} - 33 x + 63

x = \frac{21}{4}, x = 3

Solution steps

0 = 4 x^{2} - 33 x + 63

Switch sides

4 x^{2} - 33 x + 63 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 33 ) \pm ( - 33 ) ^{2} - 4 \cdot 4 \cdot 63}{2 \cdot 4}

(- 33)^{2} - 4 \cdot 4 \cdot 63 = 9

x_{1, 2} = \frac{- ( - 33 ) \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 33 ) + 9}{2 \cdot 4}, x_{2} = \frac{- ( - 33 ) - 9}{2 \cdot 4}

x = \frac{- ( - 33 ) + 9}{2 \cdot 4} : \frac{21}{4}

x = \frac{- ( - 33 ) - 9}{2 \cdot 4} : 3

The solutions to the quadratic equation are:

x = \frac{21}{4}, x = 3

(\frac{3}{x})^{3}

9 t^{2} - 18 t + 9 = 0

5 (w - 2) + 6 = 3 w - 14

7 a^{2} + 32 = 7 - 40 a

\frac{y ^{4}}{y ^{- 7}}

What are the solutions to the equation 0=4x^2-33x+63 ?
The solutions to the equation 0=4x^2-33x+63 are x= 21/4 ,x=3
Find the zeros of 0=4x^2-33x+63
The zeros of 0=4x^2-33x+63 are x= 21/4 ,x=3