What are the solutions to the equation 10x^2-6=9x ?

The solutions to the equation 10x^2-6=9x are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)

Find the zeros of 10x^2-6=9x

The zeros of 10x^2-6=9x are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)

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10 x^{2} - 6 = 9 x

x = \frac{9 + 321}{20}, x = \frac{9 - 321}{20}

Decimal Notation

x = 1.34582 \dots, x = - 0.44582 \dots

Solution steps

10 x^{2} - 6 = 9 x

Move

9 x

to the left side

10 x^{2} - 6 - 9 x = 0

Write in the standard form

a x^{2} + b x + c = 0

10 x^{2} - 9 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 10 ( - 6 )}{2 \cdot 10}

(- 9)^{2} - 4 \cdot 10 (- 6) = 321

x_{1, 2} = \frac{- ( - 9 ) \pm 321}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 321}{2 \cdot 10}, x_{2} = \frac{- ( - 9 ) - 321}{2 \cdot 10}

x = \frac{- ( - 9 ) + 321}{2 \cdot 10} : \frac{9 + 321}{20}

x = \frac{- ( - 9 ) - 321}{2 \cdot 10} : \frac{9 - 321}{20}

The solutions to the quadratic equation are:

x = \frac{9 + 321}{20}, x = \frac{9 - 321}{20}

5 x^{5} y^{2} - 80 x y^{2}

\frac{1}{3 + i}

\frac{2 ^{16}}{2 ^{4}}

t^{3} + 8

4 a^{(3)} x - 4 a^{(2)} b + 3 bm - 3 am x

What are the solutions to the equation 10x^2-6=9x ?
The solutions to the equation 10x^2-6=9x are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)
Find the zeros of 10x^2-6=9x
The zeros of 10x^2-6=9x are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)