Question

formula method
$6 {y }^{2} + 7y - 10 = 0$
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Answer 1

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$\sf \: roots \: are \: \frac{5}{6} \: and \: - 2 \\ \\ \sf \: see \: the \: attachment \: for \: step \: by \: step \: explantion \\ \sf hope \: it \: helps$

Answer 2

$\large\bf\underline\red{Answer \: :-}$

Given Quadratic equation :

$\bigstar \: \: \bf\purple{6y {}^{2} + 7y - 10 = 0}$

First know about general formula of a quadratic equation,

General formula : ax² + bx + c = 0

We multiple a and like we get, ac ( easy to solve ) And we factorize.

How to factorize equation

When we ac see the equation and think about it we have those 2 factor we multiply them we get ac, and we add them we get b.

According to the all concept,

Multiple a and c

$\dashrightarrow\sf{6\: \times \:( - 10) } \\ \dashrightarrow\sf{ - 60 } \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Factorization :

factors of -60

12 and -5

Verify :

Multiple of factor = ac

12 × -5 = -60

Addition if the factors = b

12 + (-5) = 7

Now,

$\implies\sf{y {}^{2} + 7y - 60 = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\sf{y {}^{2} + 12y - 5y - 60 = 0 } \: \: \: \: \: \: \\ \\ \implies\sf{y(y + 12) - 5(y + 12) = 0} \\ \\ \implies\sf{(y + 12)(y - 5) = 0} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\sf{y + 12 = 0 \: \: or \: \: y - 5 = 0} \: \: \\ \\ \implies\bf \red{y = - 12 \: \: \: or \: \: \: y = 5} \: \: \: \: \: \: \: \:$

Roots :

y = -12

y = 5

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