# Solve. 2x^2 + 20x = –38?

–5 ± sqrt14

It is possible to finish the squares for the issue as opposed to factoring the appearance.
2x²/2 + 20x/2 = -38/2
x² + 10x = -19
x² + 10x + 25 = 25 – 19
(x + 5)² = 6
After that, by square-root home..
√(x + 5)² = ±√6
x + 5 = ±√6
Thus, x = -5 ± √6 [Third Choice]
I really hope this can help!

polishing from the sq. 2x² + 20x = -38 x² + 10x = – 19 [ Divide via worry-unfastened factor of two] x² + 10x + 25 = -19 + 25 [upload 25 to the two the climate so as that L.H.S takes one among those x² + 2xy + y²] (x+5)² = 6 [x² + 2xy + y² = (x+y)²] x + 5 = ± ?6 [Take sq. root of the two factors , R.H.S could be ± ?6 on the grounds that (?6)² and (-?6)² are the two equivalent to six] this is exactly why x= -5 + ?6 or x = -5 – ?6 Quadratic equation origins for the equation 2x² + 20x = – 38 2x² + 20x + 38 = 0 [Taking -38 to the L.H.S] origins tend to be -(b ±?(b² – 4ac) )/ 2a the area a= 2 b = 20 and c = 38 = ( -20 ±?( (20)² – 4(2)(38) / 2(2) ) = ( -20 ±?( 4 hundred – 304) / 4) = ( ( -20 ±?ninety six / 4) = ( -20 ±?sixteen * ?6) / 4) = (-20 ±4?6)/4 = -5 ±?6 ie : -5 + ?6 and -5 – ?6 will be the origins Cheers !!!

2x^2 + 20x +38=0
2(x^2 + 20x +38)=0
x^2 + 10x + 19=0
after that factorize to perform

… 2x^2 + 20x = -38
or x^2 + 10x = -19
or x^2 + 2(x)(5) = -19
or x^2 + 2(x)(5) + 5^2 = 5^2 -19
or (x + 5)^2 = 6
or x + 5 = ± √6
or x = – 5 ± √6