What are the solutions to the equation 11x^2-98=-11+9x ?

The solutions to the equation 11x^2-98=-11+9x are x=(9+sqrt(3909))/(22),x=(9-sqrt(3909))/(22)

Find the zeros of 11x^2-98=-11+9x

The zeros of 11x^2-98=-11+9x are x=(9+sqrt(3909))/(22),x=(9-sqrt(3909))/(22)

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11 x^{2} - 98 = - 11 + 9 x

x = \frac{9 + 3909}{22}, x = \frac{9 - 3909}{22}

Decimal Notation

x = 3.25099 \dots, x = - 2.43281 \dots

Solution steps

11 x^{2} - 98 = - 11 + 9 x

Move

9 x

to the left side

11 x^{2} - 98 - 9 x = - 11

Move

11

to the left side

11 x^{2} - 9 x - 87 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 11 ( - 87 )}{2 \cdot 11}

(- 9)^{2} - 4 \cdot 11 (- 87) = 3909

x_{1, 2} = \frac{- ( - 9 ) \pm 3909}{2 \cdot 11}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3909}{2 \cdot 11}, x_{2} = \frac{- ( - 9 ) - 3909}{2 \cdot 11}

x = \frac{- ( - 9 ) + 3909}{2 \cdot 11} : \frac{9 + 3909}{22}

x = \frac{- ( - 9 ) - 3909}{2 \cdot 11} : \frac{9 - 3909}{22}

The solutions to the quadratic equation are:

x = \frac{9 + 3909}{22}, x = \frac{9 - 3909}{22}

q^{2} + 2 q - 15 = - 3 q

a \cdot (x^{2}) + b \cdot x = - c

5 t^{2} + 7 t - 1 = 0

y, y^{2} - x^{2} + 1 = 50

12 + y^{2} - 6 y = 0

What are the solutions to the equation 11x^2-98=-11+9x ?
The solutions to the equation 11x^2-98=-11+9x are x=(9+sqrt(3909))/(22),x=(9-sqrt(3909))/(22)
Find the zeros of 11x^2-98=-11+9x
The zeros of 11x^2-98=-11+9x are x=(9+sqrt(3909))/(22),x=(9-sqrt(3909))/(22)