Please solve (x+3)(x+4)=20 for x
Expand the double bracket (x+3)(x+4) first: 1) (x)(x)= x^2, 2) (x)(4)=4x ,3) (3)(x)=3x, 4) (3)(4)=12write expanded equation:x^2+4x+3x+12=20simplify and equate the right hand side to zero:x^2+7x-8=0solve for x by factorising the quadratic equation:(x+8)(x-1)=0so the solutions for x are:x=-8 and x=1
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