solve for m: 5m^2+21m+134=0
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Thanks but would you please give the exact answer ....upto this step i solved it
Answered by
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Hey dear,
● Answer -
m = (-21 + √-2239) / 10 or m = (-21 - √-2239) / 10
● Explanation -
Let's see the equation,
5m^2 + 21m + 134 = 0
Comparing with standard quadratic equation am^2 + bx + c = 0,
a = 5, b = 21, c = 134 .
Let's find value of delta,
∆ = b^2 - 4ac
∆ = 21^2 - 4×5×134
∆ = -2239
Value of x is calculated by -
m = (-b ± √∆) / 2a
m = (-21 ± √-2239) / (2×5)
m = (-21 ± √-2239) / 10
Therefore, value of m can be (-21+√-2239)/10 or (-21-√-2239)/10.
Hope this helps you.
● Answer -
m = (-21 + √-2239) / 10 or m = (-21 - √-2239) / 10
● Explanation -
Let's see the equation,
5m^2 + 21m + 134 = 0
Comparing with standard quadratic equation am^2 + bx + c = 0,
a = 5, b = 21, c = 134 .
Let's find value of delta,
∆ = b^2 - 4ac
∆ = 21^2 - 4×5×134
∆ = -2239
Value of x is calculated by -
m = (-b ± √∆) / 2a
m = (-21 ± √-2239) / (2×5)
m = (-21 ± √-2239) / 10
Therefore, value of m can be (-21+√-2239)/10 or (-21-√-2239)/10.
Hope this helps you.
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