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Question
Solve the following quadratic equation by formula method
3y2 – 20y – 23 = 0
Solution
a = 3, b = – 20, c = – 23
x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`
= `(20 ± sqrt(400 - 4(3) (- 23)))/(6)`
= `(20 ± sqrt(400 + 276))/(6)`
= `(20 ± sqrt(676))/(6)`
= `(20 ± 26)/(6)`
= `(2(10 ± 13))/(6)`
= `((10 ± 13))/(3)`
= `(10 + 13)/(3)` or `(10 - 13)/(3)`
= `(23)/(3)` or `(-3)/(3) = (23)/(3)` or – 1
The solution set is – 1 and `(23)/(5)`
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