What are the solutions to the equation x^2-3x+(-9/4)=12 ?

The solutions to the equation x^2-3x+(-9/4)=12 are x=(3+sqrt(66))/2 ,x=(3-sqrt(66))/2

Find the zeros of x^2-3x+(-9/4)=12

The zeros of x^2-3x+(-9/4)=12 are x=(3+sqrt(66))/2 ,x=(3-sqrt(66))/2

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x^{2} - 3 x + (- \frac{9}{4}) = 12

x = \frac{3 + 66}{2}, x = \frac{3 - 66}{2}

Decimal Notation

x = 5.56201 \dots, x = - 2.56201 \dots

Solution steps

x^{2} - 3 x + (- \frac{9}{4}) = 12

Refine

x^{2} - 3 x - \frac{9}{4} = 12

Move

12

to the left side

x^{2} - 3 x - \frac{57}{4} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - \frac{57}{4} )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- \frac{57}{4}) = 66

x_{1, 2} = \frac{- ( - 3 ) \pm 66}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 66}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 66}{2 \cdot 1}

x = \frac{- ( - 3 ) + 66}{2 \cdot 1} : \frac{3 + 66}{2}

x = \frac{- ( - 3 ) - 66}{2 \cdot 1} : \frac{3 - 66}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 66}{2}, x = \frac{3 - 66}{2}

x^{2} - 16 x - 64 = 0

9 x^{2} + 15 x + 6 = 0

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

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

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

What are the solutions to the equation x^2-3x+(-9/4)=12 ?
The solutions to the equation x^2-3x+(-9/4)=12 are x=(3+sqrt(66))/2 ,x=(3-sqrt(66))/2
Find the zeros of x^2-3x+(-9/4)=12
The zeros of x^2-3x+(-9/4)=12 are x=(3+sqrt(66))/2 ,x=(3-sqrt(66))/2