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xxMikexx [17]
3 years ago
6

Find the solutions of the quadratic equation 14x^2+9x+10=014x

Mathematics
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

Option B. x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

14x^{2}+9x+10=0

so

a=14\\b=9\\c=10

substitute in the formula

x=\frac{-9(+/-)\sqrt{9^{2}-4(14)(10)}} {2(14)}

x=\frac{-9(+/-)\sqrt{-479}} {28}

Remember that

i=\sqrt{-1}

substitute

x=\frac{-9(+/-)\sqrt{479}i} {28}  

x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

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