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Mathematics
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Using the Quadratics Formula
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\[2x ^{2}+7x+3=40\]
You have to first put it in the form: \(\Large ax^2+bx+c=0\) So just subtract the 40 from both sides. Then plug a, b and c into the QF: \[\Large x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
\[\frac{ x= -7 \pm \sqrt{7^{2}-4 X2X-37} }{ 4 }\]
I think what you mean there is: \(\Large x=\dfrac{ -7 \pm \sqrt{7^{2}-4 (2)(-37)} }{ 4 }\) right?? If so, then yes, that is correct so far.
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