What are the solutions to the equation 2(2a-5)^2-9(2a-5)+10=0 ?

The solutions to the equation 2(2a-5)^2-9(2a-5)+10=0 are a= 15/4 ,a= 7/2

Find the zeros of 2(2a-5)^2-9(2a-5)+10=0

The zeros of 2(2a-5)^2-9(2a-5)+10=0 are a= 15/4 ,a= 7/2

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2 (2 a - 5)^{2} - 9 (2 a - 5) + 10 = 0

a = \frac{15}{4}, a = \frac{7}{2}

Decimal Notation

a = 3.75, a = 3.5

Solution steps

2 (2 a - 5)^{2} - 9 (2 a - 5) + 10 = 0

Expand

2 (2 a - 5)^{2} - 9 (2 a - 5) + 10 : 8 a^{2} - 58 a + 105

8 a^{2} - 58 a + 105 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 58 ) \pm ( - 58 ) ^{2} - 4 \cdot 8 \cdot 105}{2 \cdot 8}

(- 58)^{2} - 4 \cdot 8 \cdot 105 = 2

a_{1, 2} = \frac{- ( - 58 ) \pm 2}{2 \cdot 8}

Separate the solutions

a_{1} = \frac{- ( - 58 ) + 2}{2 \cdot 8}, a_{2} = \frac{- ( - 58 ) - 2}{2 \cdot 8}

a = \frac{- ( - 58 ) + 2}{2 \cdot 8} : \frac{15}{4}

a = \frac{- ( - 58 ) - 2}{2 \cdot 8} : \frac{7}{2}

The solutions to the quadratic equation are:

a = \frac{15}{4}, a = \frac{7}{2}

- 6 - (- 11)

\frac{2 e ^{x} - e ^{2 x}}{2}

x^{3} \leq x^{2}

62 5^{\frac{5}{4}}

8 x^{2} + 56 x = 0

What are the solutions to the equation 2(2a-5)^2-9(2a-5)+10=0 ?
The solutions to the equation 2(2a-5)^2-9(2a-5)+10=0 are a= 15/4 ,a= 7/2
Find the zeros of 2(2a-5)^2-9(2a-5)+10=0
The zeros of 2(2a-5)^2-9(2a-5)+10=0 are a= 15/4 ,a= 7/2