Menu Close

2x-2-5x-7-0-




Question Number 85555 by luxlavanish last updated on 22/Mar/20
2x^2 +5x+7=0
$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\mathrm{0} \\ $$
Answered by MJS last updated on 23/Mar/20
ax^2 +bx+c=0  ⇒  x=((−b±(√(b^2 −4ac)))/(2a))  in this case  a=2; b=5; c=7  x=((−5±(√(5^2 −4×2×7)))/(2×2))=((−5±i(√(31)))/4)
$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${a}=\mathrm{2};\:{b}=\mathrm{5};\:{c}=\mathrm{7} \\ $$$${x}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{4}×\mathrm{2}×\mathrm{7}}}{\mathrm{2}×\mathrm{2}}=\frac{−\mathrm{5}\pm\mathrm{i}\sqrt{\mathrm{31}}}{\mathrm{4}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *