What are the solutions to the equation (x^2}{15}-\frac{2x-5)/3 = 4/5 ?

The solutions to the equation (x^2}{15}-\frac{2x-5)/3 = 4/5 are x=5+2sqrt(3),x=5-2sqrt(3)

Find the zeros of (x^2}{15}-\frac{2x-5)/3 = 4/5

The zeros of (x^2}{15}-\frac{2x-5)/3 = 4/5 are x=5+2sqrt(3),x=5-2sqrt(3)

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\frac{x ^{2}}{15} - \frac{2 x - 5}{3} = \frac{4}{5}

x = 5 + 23, x = 5 - 23

Decimal Notation

x = 8.46410 \dots, x = 1.53589 \dots

Solution steps

\frac{x ^{2}}{15} - \frac{2 x - 5}{3} = \frac{4}{5}

Find Least Common Multiplier of

15, 3, 5 : 15

Multiply by LCM=

15

\frac{x ^{2}}{15} \cdot 15 - \frac{2 x - 5}{3} \cdot 15 = \frac{4}{5} \cdot 15

Simplify

x^{2} - 5 (2 x - 5) = 12

Expand

x^{2} - 5 (2 x - 5) : x^{2} - 10 x + 25

x^{2} - 10 x + 25 = 12

Move

12

to the left side

x^{2} - 10 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 13 = 43

x_{1, 2} = \frac{- ( - 10 ) \pm 4 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 4 3}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 4 3}{2 \cdot 1}

x = \frac{- ( - 10 ) + 4 3}{2 \cdot 1} : 5 + 23

x = \frac{- ( - 10 ) - 4 3}{2 \cdot 1} : 5 - 23

The solutions to the quadratic equation are:

x = 5 + 23, x = 5 - 23

4 t^{2} - 11 t + 4 = 0

7 x^{2} + 4 x = 0

5 + 3 c^{2} = 17

24 π r - \frac{9 π}{2} r^{2} = 0

q^{2} - q - 10 = 2 q

What are the solutions to the equation (x^2}{15}-\frac{2x-5)/3 = 4/5 ?
The solutions to the equation (x^2}{15}-\frac{2x-5)/3 = 4/5 are x=5+2sqrt(3),x=5-2sqrt(3)
Find the zeros of (x^2}{15}-\frac{2x-5)/3 = 4/5
The zeros of (x^2}{15}-\frac{2x-5)/3 = 4/5 are x=5+2sqrt(3),x=5-2sqrt(3)