What are the solutions to the equation 4x^2-bx+9=0 ?

The solutions to the equation 4x^2-bx+9=0 are x=(b+sqrt(b^2-144))/8 ,x=(b-sqrt(b^2-144))/8

Find the zeros of 4x^2-bx+9=0

The zeros of 4x^2-bx+9=0 are x=(b+sqrt(b^2-144))/8 ,x=(b-sqrt(b^2-144))/8

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4 x^{2} - b x + 9 = 0

x = \frac{b + b ^{2} - 144}{8}, x = \frac{b - b ^{2} - 144}{8}

Solution steps

4 x^{2} - b x + 9 = 0

Solve with the quadratic formula

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

Simplify

(- b)^{2} - 4 \cdot 4 \cdot 9 : b^{2} - 144

x_{1, 2} = \frac{- ( - b ) \pm b ^{2} - 144}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - b ) + b ^{2} - 144}{2 \cdot 4}, x_{2} = \frac{- ( - b ) - b ^{2} - 144}{2 \cdot 4}

x = \frac{- ( - b ) + b ^{2} - 144}{2 \cdot 4} : \frac{b + b ^{2} - 144}{8}

x = \frac{- ( - b ) - b ^{2} - 144}{2 \cdot 4} : \frac{b - b ^{2} - 144}{8}

The solutions to the quadratic equation are:

x = \frac{b + b ^{2} - 144}{8}, x = \frac{b - b ^{2} - 144}{8}

8 x^{2} + 14 x = 0

0.2 x^{2} - 1.1 x - 1 = 0

n^{2} + 10 = 7 n

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

15 a x^{2} + 6 x - 3 a x = 0

What are the solutions to the equation 4x^2-bx+9=0 ?
The solutions to the equation 4x^2-bx+9=0 are x=(b+sqrt(b^2-144))/8 ,x=(b-sqrt(b^2-144))/8
Find the zeros of 4x^2-bx+9=0
The zeros of 4x^2-bx+9=0 are x=(b+sqrt(b^2-144))/8 ,x=(b-sqrt(b^2-144))/8