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3x^2-40=-7x
We move all terms to the left:
3x^2-40-(-7x)=0
We get rid of parentheses
3x^2+7x-40=0
a = 3; b = 7; c = -40;
Δ = b2-4ac
Δ = 72-4·3·(-40)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*3}=\frac{-30}{6} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*3}=\frac{16}{6} =2+2/3 $
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